OP Malhotra Class 9 Maths Solutions Chapter 11 Rectilinear Figures Exercise 11 (B)

Question 1.
(i) If in a parallelogram, the diagonals are equal in length and intersect at right angles, prove that the parallelogram will be a square.
(ii) Prove that in a rectangle, the diagonals are of equal lengths.
Answer:
(i) Given: ABCD is a parallelogram where diagonal AC = diagonal BD, and AC and BD intersect at O at right angles.

To prove: ABCD is a square.
Proof:
Consider triangles \( \triangle ABC \) and \( \triangle ADB \).
\( AB = AB \) (This side is common to both triangles).
\( AC = BD \) (This is given in the problem).
\( BC = AD \) (These are opposite sides of a parallelogram).
Therefore, \( \triangle ABC \cong \triangle ADC \) (by SSS axiom).
So, \( \angle ABC = \angle BAD \) (by c.p.c.t.).
We know that \( \angle ABC + \angle BAD = 180^\circ \) (These are co-interior angles, as AB is parallel to CD and AD is a transversal).
Since \( \angle ABC = \angle BAD \), we can write \( \angle ABC + \angle ABC = 180^\circ \).
\( 2 \angle ABC = 180^\circ \)
\( \implies \angle ABC = 90^\circ \).
This means \( \angle ABC = \angle BAD = 90^\circ \).
Now, let's look at triangles \( \triangle AOB \) and \( \triangle BOC \).
\( \angle AOB = \angle BOC \) (Each of these angles is \( 90^\circ \) because the diagonals intersect at right angles).
\( BO = BO \) (This side is common to both triangles).
\( AO = OC \) (Diagonals of a parallelogram bisect each other, which means they cut each other exactly in half).
Therefore, \( \triangle AOB \cong \triangle BOC \) (by SAS axiom).
So, \( AB = BC \) (by c.p.c.t.).
We also know that \( AB = CD \) and \( BC = AD \) (These are opposite sides of a parallelogram).
Putting all these equalities together, we get \( AB = BC = CD = DA \).
Since all sides are equal and each angle is \( 90^\circ \), ABCD is a square. This completes the proof.

(ii) Given: ABCD is a rectangle, and AC and BD are its diagonals.

To prove: \( AC = BD \).
Proof:
Consider triangles \( \triangle ABC \) and \( \triangle BAD \).
\( AB = AB \) (This side is common to both triangles).
\( BC = AD \) (These are opposite sides of a rectangle, which means they are equal).
\( \angle ABC = \angle BAD \) (Each angle in a rectangle is \( 90^\circ \)).
Therefore, \( \triangle ABC \cong \triangle ADB \) (by SAS axiom).
So, \( AC = BD \) (by c.p.c.t.).
This proves that the diagonals of a rectangle are equal in length.
In simple words: For part (i), if a parallelogram has diagonals that are equal and cross at a right angle, it must be a square because all its sides will be equal and all angles will be 90 degrees. For part (ii), in a rectangle, the two main lines (diagonals) from corner to opposite corner are always the same length because of how rectangles are shaped.

🎯 Exam Tip: When proving properties of quadrilaterals, always clearly state your given information, what you need to prove, and the geometric reasons (axioms or properties) for each step of your proof.

Question 2. In the figure, ABCD is a parallelogram. M is the mid-point of AC; X, Y are points on AB and DC respectively such that AX = CY.
Prove that:
(i) \( \triangle AXM = \triangle CYM \)
(ii) XMY is a straight line.
Answer:

Given: In parallelogram ABCD, AC is its diagonal. X is a point on AB and Y is a point on CD, such that AX = CY. M is the mid-point of AC.
To prove: (i) \( \triangle AXM = \triangle CYM \) (ii) XMY is a straight line.
Construction: Join XM and YM.
Proof:
(i) Consider triangles \( \triangle AMX \) and \( \triangle CMY \).
\( AX = CY \) (This is given in the problem).
\( AM = CM \) (M is the mid-point of AC, so it divides AC into two equal parts).
\( \angle MAX = \angle YCM \) (These are alternate interior angles because AB is parallel to DC, and AC is a transversal).
Therefore, \( \triangle AMX \cong \triangle CMY \) (by SAS axiom).

(ii) From the congruence in part (i), we know that corresponding parts of congruent triangles are equal.
So, \( \angle AMX = \angle CMY \) (by c.p.c.t.).
Angles \( \angle AMX \) and \( \angle CMY \) are vertically opposite angles.
When two line segments meet and the vertically opposite angles formed are equal, it means the line segments are part of a single straight line. Thus, XMY forms a straight line.
In simple words: For part (i), we can show two small triangles, AXM and CYM, are exactly the same size and shape by comparing their sides and angles. This is because M cuts the diagonal AC in half, and AX equals CY. For part (ii), because these two triangles are the same, the angles at M are vertically opposite and equal, which means the line XMY must be a single straight line.

🎯 Exam Tip: When proving triangles congruent, clearly state the three conditions (Side-Side-Side, Side-Angle-Side, Angle-Side-Angle, Angle-Angle-Side, or Right angle-Hypotenuse-Side) and the conclusion (CPCTC - Corresponding Parts of Congruent Triangles are Congruent) to justify equal angles or sides.

Question 3. Prove that:
(i) A diagonal of a square makes an angle of \( 45^\circ \) with the side of the square.
(ii) The diagonals of a rhombus are at right angles.
(iii) A diagonal of a rhombus bisects the angles at vertices.
Answer:
(i) Given: ABCD is a square, and BD is its diagonal.

To prove: \( \angle ABD = 45^\circ \).
Proof:
In a square, all sides are equal, so \( AB = AD \).
Since two sides of \( \triangle ABD \) are equal, it is an isosceles triangle.
In \( \triangle ABD \), the angles opposite to the equal sides are also equal, so \( \angle ABD = \angle ADB \).
We know that \( \angle DAB = 90^\circ \) (Each angle of a square is \( 90^\circ \)).
The sum of angles in a triangle is \( 180^\circ \), so in \( \triangle ABD \):
\( \angle ABD + \angle ADB + \angle DAB = 180^\circ \)
Since \( \angle ABD = \angle ADB \), we have:
\( 2 \angle ABD + 90^\circ = 180^\circ \)
\( 2 \angle ABD = 180^\circ - 90^\circ \)
\( 2 \angle ABD = 90^\circ \)
\( \implies \angle ABD = 45^\circ \).
This proves that a diagonal of a square makes an angle of \( 45^\circ \) with the side of the square.

(ii) Given: In rhombus ABCD, its diagonals AC and BD bisect each other at O.

To prove: \( \angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ \).
Proof:
Consider triangles \( \triangle AOB \) and \( \triangle COB \).
\( AO = OC \) (Diagonals of a parallelogram bisect each other, and a rhombus is a special parallelogram).
\( BO = BO \) (This side is common to both triangles).
\( AB = BC \) (All sides of a rhombus are equal).
Therefore, \( \triangle AOB \cong \triangle COB \) (by SSS axiom).
So, \( \angle AOB = \angle COB \) (by c.p.c.t.).
We know that \( \angle AOB + \angle COB = 180^\circ \) (These angles form a linear pair on the straight line AC).
Since \( \angle AOB = \angle COB \), we have \( 2 \angle AOB = 180^\circ \).
\( \implies \angle AOB = 90^\circ \).
Because \( \angle AOB = 90^\circ \), it means \( \angle COB \) is also \( 90^\circ \).
Similarly, we can prove that \( \angle COD = 90^\circ \) and \( \angle DOA = 90^\circ \).
Hence, the diagonals of a rhombus bisect each other at right angles.

(iii) Given: In rhombus ABCD, AC is its diagonal.

To prove: AC bisects \( \angle A \) and \( \angle C \).
Proof:
Consider triangles \( \triangle ABC \) and \( \triangle ADC \).
\( AB = CD \) (These are opposite sides of a rhombus, so they are equal).
\( BC = AD \) (These are opposite sides of a rhombus, so they are equal).
\( AC = AC \) (This side is common to both triangles).
Therefore, \( \triangle ABC \cong \triangle ADC \) (by SSS axiom).
So, \( \angle CAB = \angle CAD \) (by c.p.c.t.). This means AC bisects \( \angle A \).
Also, \( \angle ACB = \angle ACD \) (by c.p.c.t.). This means AC bisects \( \angle C \).
Hence, a diagonal of a rhombus bisects the angles at the vertices it connects.
In simple words: For a square, the line that cuts it in half diagonally always makes a 45-degree angle with the sides. For a rhombus, its two diagonals always cross each other at a perfect 90-degree angle. Also, these diagonals cut the corner angles of the rhombus exactly in half.

🎯 Exam Tip: Remember the specific properties of squares and rhombuses. A square is a special type of rhombus and a rectangle, so it has properties of both. A rhombus has all sides equal, and its diagonals are perpendicular bisectors of each other and bisect the vertex angles.

Question 4. In the figure, PQRS is a parallelogram. QU and ST are perpendicular on the diagonal PR.
Prove that
(i) \( \triangle STR \cong \triangle QUP \)
(ii) ST = QU
Answer:

Given: PQRS is a parallelogram in which PR is its diagonal. ST is perpendicular to PR, and QU is perpendicular to PR.
To prove: (i) \( \triangle STR \cong \triangle QUP \) (ii) ST = QU.
Proof:
(i) Consider triangles \( \triangle STR \) and \( \triangle QUP \).
\( SR = PQ \) (These are opposite sides of a parallelogram, so they are equal in length).
\( \angle STR = \angle QUP \) (Both angles are \( 90^\circ \) because ST and QU are perpendicular to PR).
\( \angle SRT = \angle QPU \) (These are alternate interior angles because SR is parallel to PQ, and PR is a transversal cutting them).
Therefore, \( \triangle STR \cong \triangle QUP \) (by AAS axiom, which means Angle-Angle-Side).

(ii) Since \( \triangle STR \cong \triangle QUP \), their corresponding parts are equal.
So, \( ST = QU \) (by c.p.c.t.).
In simple words: In this parallelogram, we have two lines, ST and QU, that drop straight down (perpendicular) to the diagonal PR. We can show that the two triangles formed, STR and QUP, are exactly the same size and shape because they share angles and sides in a specific way. Because they are congruent, the lengths ST and QU must also be equal.

🎯 Exam Tip: When dealing with proofs involving parallel lines and transversals, always look for alternate interior angles, corresponding angles, or co-interior angles to establish angle equalities.

Question 5. In the figure, PQRS is a parallelogram. PO and QO are respectively the angle bisectors of \( \angle P \) and \( \angle Q \). Line LOM is drawn parallel to PQ. Prove that
(i) PL = QM
(ii) LO = OM
Answer:

Given: In parallelogram PQRS, PO bisects \( \angle P \), and QO bisects \( \angle Q \). Line LOM is drawn through O and is parallel to PQ.
To prove: (i) PL = QM (ii) LO = OM.
Proof:
Since PO is the bisector of \( \angle P \), this means \( \angle 1 = \angle 2 \).
Similarly, QO is the bisector of \( \angle Q \), so \( \angle 5 = \angle 6 \).
We are given that LM is parallel to PQ.
Consider the quadrilateral LPQM. Since PQRS is a parallelogram, PS is parallel to QR. Also, LM is parallel to PQ, and LP is part of PS, while QM is part of QR. This makes LPQM a parallelogram.
(i) In parallelogram LPQM, opposite sides are equal.
So, \( PL = QM \). This proves part (i).
(ii) Because LM is parallel to PQ, we have alternate interior angles:
\( \angle 3 = \angle 2 \) and \( \angle 4 = \angle 5 \).
Since \( \angle 1 = \angle 2 \) (PO is a bisector) and \( \angle 3 = \angle 2 \), it means \( \angle 1 = \angle 3 \).
In \( \triangle PLO \), since \( \angle 1 = \angle 3 \), the sides opposite to these equal angles must also be equal. So, \( PL = LO \).
Similarly, since \( \angle 5 = \angle 6 \) (QO is a bisector) and \( \angle 4 = \angle 5 \), it means \( \angle 4 = \angle 6 \).
In \( \triangle QMO \), since \( \angle 4 = \angle 6 \), the sides opposite to these equal angles must also be equal. So, \( QM = OM \).
From part (i), we proved \( PL = QM \).
Since \( PL = LO \) and \( QM = OM \), and \( PL = QM \), it follows that \( LO = OM \).
This proves part (ii).
In simple words: In this parallelogram, the lines PO and QO cut angles P and Q exactly in half. A line LOM is drawn through point O, running parallel to the bottom side PQ. We show that the part PL on the left is the same length as QM on the right. Then, we prove that the point O is exactly in the middle of the line LM, making LO and OM equal in length.

🎯 Exam Tip: When angle bisectors are involved with parallel lines, look for isosceles triangles formed by alternate interior angles to prove side equalities.

Question 6.
(i) ABCD is a parallelogram. L and M are points on AB and DC respectively and AL = CM. Prove that LM and BD bisect each other.
(ii) ABCD is a parallelogram. AB is produced to E such that BE = AB. Prove that ED bisects BC.
Answer:
(i) Given: In parallelogram ABCD, L is a point on AB and M is a point on DC such that AL = CM. BD is a diagonal.

To prove: LM and BD bisect each other (i.e., BO = OD and LO = OM).
Proof:
In parallelogram ABCD, we know that opposite sides are equal, so \( AB = CD \).
We are given \( AL = CM \).
Subtracting AL from AB and CM from CD:
\( AB - AL = CD - CM \)
\( \implies LB = MD \).
Now consider triangles \( \triangle LOB \) and \( \triangle DOM \).
\( LB = MD \) (This was just proved).
\( \angle LBO = \angle MDO \) (These are alternate interior angles because AB is parallel to DC, and BD is a transversal).
\( \angle LOB = \angle DOM \) (These are vertically opposite angles).
Therefore, \( \triangle LOB \cong \triangle DOM \) (by AAS axiom).
Since the triangles are congruent, their corresponding parts are equal:
\( OB = OD \) (by c.p.c.t.).
\( OL = OM \) (by c.p.c.t.).
Hence, LM and BD bisect each other.

(ii) Given: In parallelogram ABCD, AB is extended to E such that \( BE = AB \).

To prove: ED bisects BC.
Proof:
In a parallelogram, opposite sides are equal, so \( AB = DC \).
We are given that \( AB = BE \).
From these two facts, we can say \( DC = BE \).
Now consider triangles \( \triangle BOE \) and \( \triangle COD \).
\( BE = DC \) (This was just proved).
\( \angle BOE = \angle COD \) (These are vertically opposite angles).
\( \angle EBO = \angle DCO \) (These are alternate interior angles because AB is parallel to DC, and BC is a transversal).
Therefore, \( \triangle BOE \cong \triangle COD \) (by AAS axiom).
Since the triangles are congruent, their corresponding parts are equal:
\( BO = OC \) (by c.p.c.t.).
This means that O is the mid-point of BC, so ED bisects BC.
In simple words: For part (i), if you have a parallelogram and mark points L and M such that AL equals CM, then the line connecting L and M will cut the diagonal BD exactly in half, and BD will also cut LM exactly in half. For part (ii), if you extend one side of a parallelogram (AB) to a point E so that BE is as long as AB, then the line connecting D to E will cut the side BC exactly in half.

🎯 Exam Tip: When a problem asks to prove that segments bisect each other, look for congruent triangles where the segments are corresponding parts that are cut into equal halves.

Question 7.
(i) The diagonals of a parallelogram intersect at O. A line through O intersects AB at X and DC at Y. Prove that OX = OY.
(ii) In a triangle ABC, median AD is produced to X, such that AD = DX. Prove that ABXC is a parallelogram.
Answer:
(i) Given: In parallelogram ABCD, diagonals AC and BD intersect each other at O. A line XOY is drawn which meets AB at X and CD at Y.

To prove: \( OX = OY \).
Proof:
In a parallelogram, diagonals bisect each other, so \( OB = OD \).
Consider triangles \( \triangle BOX \) and \( \triangle DOY \).
\( \angle BOX = \angle DOY \) (These are vertically opposite angles).
\( \angle OBX = \angle ODY \) (These are alternate interior angles because AB is parallel to DC, and BD is a transversal).
Therefore, \( \triangle XOB \cong \triangle YOD \) (by AAS axiom).
Since the triangles are congruent, their corresponding parts are equal:
\( OX = OY \) (by c.p.c.t.).
This proves that the line segment XOY passing through the intersection of diagonals is bisected by O.

(ii) Given: In \( \triangle ABC \), AD is the median. AD is produced to X such that \( DX = AD \). BX and XC are joined.

To prove: ABXC is a parallelogram.
Proof:
Consider triangles \( \triangle ADB \) and \( \triangle XDC \).
\( AD = DX \) (This is given).
\( BD = DC \) (D is the mid-point of BC because AD is a median).
\( \angle ADB = \angle XDC \) (These are vertically opposite angles).
Therefore, \( \triangle ADB \cong \triangle XDC \) (by SAS axiom).
Since the triangles are congruent, their corresponding parts are equal:
\( AB = CX \) (by c.p.c.t.).
\( \angle BAD = \angle CXD \) (by c.p.c.t.).
Since \( \angle BAD \) and \( \angle CXD \) are alternate interior angles and are equal, it means \( AB \) is parallel to \( XC \).
Now, we have a quadrilateral ABXC where \( AB = XC \) and \( AB \parallel XC \).
A quadrilateral with one pair of opposite sides that are both equal and parallel is a parallelogram.
Therefore, ABXC is a parallelogram.
In simple words: For part (i), if you draw a straight line through the center point (O) of a parallelogram's diagonals, that line will be cut in half by O. For part (ii), if you take a triangle, extend its median (a line from a corner to the middle of the opposite side) past the middle point by the same length, and then connect the new endpoint to the other corners, you will form a parallelogram.

🎯 Exam Tip: To prove a figure is a parallelogram, remember the key conditions: either both pairs of opposite sides are parallel, both pairs are equal, one pair is both equal and parallel, or diagonals bisect each other.

Question 8. In a parallelogram ABCD, the bisector of \( \angle A \) meets DC in E and \( AB = 2AD \). Prove that (i) BE bisects \( \angle B \), (ii) \( \angle AEB \) is a right angle.
Answer:

Given: In parallelogram ABCD, AE is the bisector of \( \angle A \), meeting CD at E. Also, \( AB = 2AD \). EB is joined.
To prove: (i) BE bisects \( \angle B \), (ii) \( \angle AEB = 90^\circ \).
Construction: Take F as the mid-point of AB. Join EF.
Proof:
We are given \( AB = 2AD \). Since F is the mid-point of AB, \( AF = \frac{1}{2} AB \).
So, \( AF = AD \).
Since AE is the bisector of \( \angle A \), we have \( \angle 1 = \angle 2 \).
In parallelogram ABCD, AD is parallel to BE. Also, AE is a transversal.
So, \( \angle 2 = \angle AED \) (alternate interior angles).
Since \( \angle 1 = \angle 2 \), it means \( \angle 1 = \angle AED \).
In \( \triangle ADE \), because \( \angle 1 = \angle AED \), the sides opposite to these angles are equal: \( AD = DE \).
Since \( AF = AD \) and \( AD = DE \), we have \( AF = DE \).
Also, AD is parallel to FE (from mid-point theorem or construction).
So, AFED is a parallelogram where adjacent sides \( AD = AF \). This means AFED is a rhombus.
In a parallelogram ABCD, \( AD \parallel BC \) and \( AB \parallel DC \).
Since \( E \) is on DC and \( F \) is on AB, and \( AFED \) is a rhombus, \( EF \parallel AD \).
Since \( AD \parallel BC \), it implies \( EF \parallel BC \).
Also, \( EF = AD \) (sides of rhombus). Since \( AD = BC \) (opposite sides of parallelogram), we get \( EF = BC \).
So, FBCE is a parallelogram (one pair of opposite sides equal and parallel).
Since \( AB = 2AD \), and \( AF = AD \), we have \( FB = AB - AF = 2AD - AD = AD \).
Thus, \( FB = AD = BC \). So, FBCE is a parallelogram with adjacent sides equal, meaning FBCE is also a rhombus.
(i) In rhombus FBCE, BE is a diagonal. A property of a rhombus is that its diagonals bisect the angles at the vertices. Therefore, BE bisects \( \angle B \).

(ii) We know that \( \angle A + \angle B = 180^\circ \) (These are co-interior angles in a parallelogram).
Since AE bisects \( \angle A \) and BE bisects \( \angle B \), we have:
\( \frac{1}{2} \angle A + \frac{1}{2} \angle B = \frac{1}{2} (180^\circ) \)
\( \implies \angle EAB + \angle EBA = 90^\circ \).
Now consider \( \triangle AEB \). The sum of angles in a triangle is \( 180^\circ \).
\( \angle EAB + \angle EBA + \angle AEB = 180^\circ \)
Substitute \( \angle EAB + \angle EBA = 90^\circ \):
\( 90^\circ + \angle AEB = 180^\circ \)
\( \implies \angle AEB = 180^\circ - 90^\circ \)
\( \implies \angle AEB = 90^\circ \).
Hence, \( \angle AEB \) is a right angle.
In simple words: In a parallelogram, if a line from corner A cuts angle A in half and reaches side DC at E, and side AB is twice as long as AD, then two things happen. First, the line BE cuts angle B in half. Second, the angle AEB formed by these lines is always a right angle, which means it is 90 degrees.

🎯 Exam Tip: For problems involving angle bisectors in parallelograms, always remember that adjacent angles add up to \( 180^\circ \), which simplifies to \( 90^\circ \) when bisected, often leading to a right angle in the triangle formed by the bisectors.

Question 9.
(i) In the figure, ABCD is a parallelogram. X is the mid-point of AD and Y is the mid-point of BC. Prove that AYCX is a parallelogram and that XY and BD bisect each other.
(ii) In the figure, ABCD is a parallelogram. BM bisects \( \angle ABC \), and DN bisects \( \angle ADC \). Prove that: (i) BNDM is a parallelogram, (ii) BM = DN.
(iii) In the figure, BM bisects \( \angle B \) and AN bisects \( \angle A \) of parallelogram ABCD. Prove that : (i) MN = CD (ii) ABNM is a rhombus
Answer:
(i) Given: In parallelogram ABCD, X is the mid-point of AD and Y is the mid-point of BC. BD is its diagonal. AY and CX are joined.

To prove: (i) AYCX is a parallelogram (ii) XY and BD bisect each other.
Proof:
In parallelogram ABCD, \( AD \parallel BC \) and \( AD = BC \) (opposite sides).
Since X is the mid-point of AD, \( AX = \frac{1}{2} AD \).
Since Y is the mid-point of BC, \( CY = \frac{1}{2} BC \).
Because \( AD = BC \), it follows that \( AX = CY \).
Since \( AD \parallel BC \), it also means \( AX \parallel CY \).
(i) In quadrilateral AYCX, we have \( AX = CY \) and \( AX \parallel CY \).
A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.
Therefore, AYCX is a parallelogram.

To prove that XY and BD bisect each other, we need to show that they cross at their midpoints.
Consider triangles \( \triangle XOD \) and \( \triangle BOY \).
\( XD = BY \) (Since X and Y are mid-points of equal parallel sides AD and BC, \( \frac{1}{2} AD = \frac{1}{2} BC \)).
\( \angle XDO = \angle YBO \) (These are alternate interior angles because AD is parallel to BC, and BD is a transversal).
\( \angle DOX = \angle BOY \) (These are vertically opposite angles).
Therefore, \( \triangle XOD \cong \triangle BOY \) (by AAS axiom).
Since the triangles are congruent:
\( XO = OY \) (by c.p.c.t.).
\( DO = OB \) (by c.p.c.t.).
This means O is the mid-point of both XY and BD. Hence, XY and BD bisect each other.

(ii) Given: In parallelogram ABCD, BM bisects \( \angle ABC \), and DN bisects \( \angle ADC \).

To prove: (i) BNDM is a parallelogram (ii) BM = DN.
Proof:
Since ABCD is a parallelogram, \( AD \parallel BC \) and \( AB \parallel DC \). Also, \( \angle B = \angle D \) (opposite angles).
Since BM bisects \( \angle ABC \), \( \angle ABM = \angle MBC = \frac{1}{2} \angle B \).
Since DN bisects \( \angle ADC \), \( \angle ADN = \angle NDC = \frac{1}{2} \angle D \).
Since \( \angle B = \angle D \), it follows that \( \frac{1}{2} \angle B = \frac{1}{2} \angle D \).
So, \( \angle ABM = \angle NDC \).
Also, \( AD \parallel BC \), and BM is a transversal, so \( \angle DMB = \angle MBC \) (alternate interior angles).
Similarly, \( AB \parallel DC \), and DN is a transversal, so \( \angle ADN = \angle BND \) (alternate interior angles).
(i) In quadrilateral BNDM, we have \( AD \parallel BC \), which implies \( DM \parallel BN \).
Also, \( \angle ABM = \angle NDC \). And \( \angle MBN = \angle BDM \) (alternate interior angles, as AB || DC, BD transversal).
This isn't directly proving BNDM is a parallelogram. Let's try another approach for (i).
In parallelogram ABCD, \( AD \parallel BC \).
M is on AD and N is on BC. So, \( AM \parallel BN \).
We know \( \angle D = \angle B \). Since DN and BM are angle bisectors, \( \angle CDN = \frac{1}{2} \angle D \) and \( \angle ABM = \frac{1}{2} \angle B \). Thus \( \angle CDN = \angle ABM \).
Also, \( AD \parallel BC \), so \( \angle AMB = \angle MBC \) (alternate interior angles).
Since BM bisects \( \angle B \), \( \angle ABM = \angle MBC \). So, \( \angle AMB = \angle ABM \).
This makes \( \triangle ABM \) an isosceles triangle, so \( AM = AB \).
Similarly, \( DN \) bisects \( \angle D \), so \( \angle CDN = \angle AND \) (alternate interior angles).
Since \( \angle CDN = \angle ADN \), then \( \angle ADN = \angle AND \).
This makes \( \triangle AND \) an isosceles triangle, so \( AN = AD \).
In parallelogram ABCD, \( AD = BC \). So \( AN = BC \). This contradicts N being a midpoint if we use M on AD.
Let's re-read the setup: BM bisects \( \angle B \), DN bisects \( \angle D \).
Since \( AD \parallel BC \), \( \angle AMB = \angle MBC \) (alternate interior angles).
Since BM bisects \( \angle B \), \( \angle ABM = \angle MBC \).
Therefore, \( \angle ABM = \angle AMB \), which means \( AM = AB \).
Similarly, since \( AB \parallel DC \), \( \angle DNC = \angle ABN \) (alternate interior angles). No, this is incorrect.
Let's use the property that angles in a parallelogram are equal for opposite angles.
\( \angle ADC = \angle ABC \).
Since DN bisects \( \angle ADC \), \( \angle ADN = \angle NDC = \frac{1}{2} \angle ADC \).
Since BM bisects \( \angle ABC \), \( \angle ABM = \angle MBC = \frac{1}{2} \angle ABC \).
Thus \( \angle ADN = \angle ABM \).
Also, \( \angle ADM = \angle NBC \) (since AD || BC, and DN, BM are internal lines). No, this is not a direct consequence.
Let's use the standard result: opposite sides are parallel and equal.
In parallelogram ABCD, \( AD \parallel BC \), so \( DM \parallel BN \).
Now, we need to show that \( DM = BN \).
As shown before, in \( \triangle ABM \), \( AM = AB \).
Similarly, in \( \triangle CDN \), \( CN = CD \). (Because \( \angle DNC = \angle BCN \) (alternate angles with AD || BC and DN transversal not applicable here. Let's use \( \angle CND = \angle NDC \) due to bisector and alternate interior angles).
In parallelogram ABCD, \( \angle A + \angle D = 180^\circ \).
Since \( AB \parallel DC \), \( \angle ABM = \angle BND \) (alternate interior angles, if BM and DN are lines in relation to each other). This is not correct without more construction.
Let's go back to \( AM = AB \).
We know \( AB = CD \). So \( AM = CD \).
Since \( AD = BC \), then \( DM = AD - AM = AD - CD \). This doesn't directly simplify to \( BN \).
A key property of angle bisectors in parallelograms is that they create isosceles triangles.
Since \( AD \parallel BC \), and BM is a transversal, \( \angle AMB = \angle MBC \).
Since BM is angle bisector, \( \angle ABM = \angle MBC \).
Therefore, \( \angle AMB = \angle ABM \), making \( \triangle ABM \) isosceles with \( AM = AB \).
Similarly, for \( \angle D \): \( AB \parallel DC \), and DN is a transversal. \( \angle CDN = \angle DNA \) (alternate interior angles).
Since DN bisects \( \angle D \), \( \angle CDN = \angle ADN \).
Therefore, \( \angle DNA = \angle ADN \), making \( \triangle ADN \) isosceles with \( AN = AD \).
This implies that M is on AD and N is on BC.
(i) In BNDM, \( DM = AD - AM = AD - AB \). This isn't helpful.
Let's consider the quadrilateral BNDM directly.
We have \( BN \parallel DM \) (since BC \( \parallel \) AD).
We need to show \( BM \parallel DN \).
Since \( \angle D = \angle B \), and DN and BM bisect these angles, \( \angle ABM = \angle CDN \).
Consider \( \triangle DCN \). \( CD = CN \) (since \( \angle CDN = \angle CND \) from alt. interior \( \angle AD N \) and bisector properties). So \( CN = CD = AB \).
This implies \( BN = BC - CN = AD - AB \).
In \( \triangle ABM \), \( AM = AB \). So \( DM = AD - AM = AD - AB \).
Therefore, \( BN = DM \).
Since \( BN \parallel DM \) and \( BN = DM \), quadrilateral BNDM is a parallelogram.

(ii) Since BNDM is a parallelogram (proved above), its opposite sides are equal.
Therefore, \( BM = DN \).

(iii) Given: In parallelogram ABCD, BM bisects \( \angle B \) and AN bisects \( \angle A \).

To prove: (i) MN = CD (ii) ABNM is a rhombus.
Proof:
In parallelogram ABCD, adjacent angles sum to \( 180^\circ \), so \( \angle A + \angle B = 180^\circ \).
Since AN bisects \( \angle A \) and BM bisects \( \angle B \), we can divide the sum by 2:
\( \frac{1}{2} \angle A + \frac{1}{2} \angle B = \frac{1}{2} (180^\circ) \)
\( \implies \angle OAB + \angle OBA = 90^\circ \) (where O is the intersection of AN and BM).
In \( \triangle AOB \), the sum of angles is \( 180^\circ \).
\( \angle AOB + (\angle OAB + \angle OBA) = 180^\circ \)
\( \angle AOB + 90^\circ = 180^\circ \)
\( \implies \angle AOB = 90^\circ \).
This means AN and BM are perpendicular to each other.

(i) Consider \( \triangle AOM \) and \( \triangle BON \).
\( AO = ON \) and \( BO = OM \) (This is what we need to prove if ABNM is a parallelogram).
In parallelogram ABCD, \( AD \parallel BC \).
Since AN bisects \( \angle A \), \( \angle DAN = \angle NAB \).
Also, \( \angle ANB = \angle DAN \) (alternate interior angles because AD \( \parallel \) BC).
Therefore, \( \angle ANB = \angle NAB \), which means \( \triangle ABN \) is an isosceles triangle with \( AB = BN \).
Similarly, since BM bisects \( \angle B \), \( \angle ABM = \angle MBC \).
And \( \angle AMB = \angle MBC \) (alternate interior angles because AD \( \parallel \) BC).
Therefore, \( \angle ABM = \angle AMB \), which means \( \triangle ABM \) is an isosceles triangle with \( AM = AB \).
From these, we have \( AM = AB \) and \( BN = AB \). So, \( AM = BN \).
Also, \( MN \) is part of the line AD, and \( N \) is on BC.
We know \( AB \parallel DC \) and \( AD \parallel BC \).
Consider quadrilateral ABNM. We have \( AM = AB \) and \( BN = AB \).
This means \( AM = BN \). Also, \( AM \parallel BN \) (since AD \( \parallel \) BC).
So, ABNM is a parallelogram.
Since ABNM is a parallelogram, \( MN = AB \) (opposite sides).
We know that \( AB = CD \) (opposite sides of parallelogram ABCD).
Therefore, \( MN = CD \). This proves part (i).

(ii) We have shown that ABNM is a parallelogram. We also found that \( AB = AM \) (from \( \triangle ABM \) being isosceles).
A parallelogram with adjacent sides equal is a rhombus.
Therefore, ABNM is a rhombus. This proves part (ii).
In simple words: For Q9(i), if you connect the midpoints of AD and BC in a parallelogram to the opposite corners, you form another parallelogram. Also, the line connecting these midpoints and the diagonal BD will cut each other exactly in half. For Q9(ii), if two lines (BM and DN) cut angles B and D in half in a parallelogram, the shape BNDM formed inside will also be a parallelogram, and the lengths BM and DN will be equal. For Q9(iii), if lines BM and AN cut angles B and A in half, then the line segment MN will be the same length as side CD, and the inner shape ABNM will be a rhombus (a diamond shape with all equal sides).

🎯 Exam Tip: When angle bisectors are drawn from adjacent vertices in a parallelogram, they create an angle of 90 degrees at their intersection. Also, if a parallelogram has an adjacent pair of sides equal, it is a rhombus.

Question 10. A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B; the four bisectors form a quadrilateral ACBD. Prove that:
(i) ACBD is a rectangle.
(ii) CD is parallel to the original parallel lines.

Answer:
(i) Given that PQ and RS are two parallel lines cut by a transversal LM. The bisectors of the interior angles at A and B form a quadrilateral ACBD. In parallel lines PQ and RS, with transversal LM, we know that the sum of consecutive interior angles is 180 degrees.
So, \( \angle PAB + \angle RBA = 180^\circ \).
AC bisects \( \angle PAB \), so \( \angle CAB = \frac{1}{2} \angle PAB \).
BC bisects \( \angle RBA \), so \( \angle CBA = \frac{1}{2} \angle RBA \).
In triangle ABC, the sum of angles \( \angle CAB + \angle CBA = \frac{1}{2} (\angle PAB + \angle RBA) = \frac{1}{2} (180^\circ) = 90^\circ \).
Therefore, \( \angle ACB = 180^\circ - (\angle CAB + \angle CBA) = 180^\circ - 90^\circ = 90^\circ \).
Similarly, for the other pair of interior angles, \( \angle QAB + \angle SBA = 180^\circ \).
AD bisects \( \angle QAB \), so \( \angle DAB = \frac{1}{2} \angle QAB \).
BD bisects \( \angle SBA \), so \( \angle DBA = \frac{1}{2} \angle SBA \).
In triangle ABD, the sum of angles \( \angle DAB + \angle DBA = \frac{1}{2} (\angle QAB + \angle SBA) = \frac{1}{2} (180^\circ) = 90^\circ \).
Therefore, \( \angle ADB = 180^\circ - (\angle DAB + \angle DBA) = 180^\circ - 90^\circ = 90^\circ \).
If AC || BD and AD || BC (which can be derived from the properties of angle bisectors and parallel lines), then ACBD is a parallelogram. Since one angle, \( \angle ACB \), is 90 degrees, ACBD is a rectangle. A rectangle has all its angles as right angles.
(ii) For CD to be parallel to PQ or RS, we use the properties of the rectangle ACBD. The diagonals of a rectangle bisect each other and are equal in length. Let O be the intersection of diagonals AB and CD.
So, \( OA = OC \). This means triangle OAC is an isosceles triangle.
Thus, \( \angle OAC = \angle OCA \).
We know that AC is the bisector of \( \angle PAB \). So \( \angle CAB \) is the same as \( \angle OAC \).
Since PQ is a line, \( \angle CAB \) and \( \angle OCA \) are alternate interior angles if CD is parallel to PQ.
If CD || PQ, then \( \angle OCA = \angle CAB \) (alternate interior angles).
Since \( \angle OAC = \angle OCA \) is true, and \( \angle CAB = \angle OAC \), then \( \angle CAB = \angle OCA \) which implies CD || PQ or RS. This demonstrates the parallelism.
In simple words: When two parallel lines are crossed by another line, and you draw lines that cut the angles in half inside, the shape formed by these cutting lines will be a rectangle. Also, one side of this rectangle (CD) will always be parallel to the original two parallel lines.

🎯 Exam Tip: Remember that the sum of interior angles on the same side of a transversal between parallel lines is 180°. Bisecting them leads to a 90° angle in the triangle formed by the bisectors.

Question 11. In the figures,
(i) If in a parallelogram, the diagonals are equal in length and intersect at right angles, prove that the parallelogram will be a square.
(ii) Prove that in a rectangle, the diagonals are of equal lengths.
(iii) ABCD is a parallelogram. ABML and ADXY are squares. Prove that ACXM is an isosceles triangle.
(iv) ABCD and ALMN are squares. Prove that BL = DN.

Answer:
(i) In parallelogram ABCD, if diagonals AC and BD are equal and intersect at right angles at O, we need to prove it's a square.

Given: ABCD is a parallelogram with \( AC = BD \) and \( AC \perp BD \) at O.
To prove: ABCD is a square.
Proof:
In \( \triangle AOB \) and \( \triangle BOC \):
\( AO = OC \) (Diagonals of a parallelogram bisect each other).
\( BO = BO \) (Common side).
\( \angle AOB = \angle BOC = 90^\circ \) (Given, diagonals intersect at right angles).
So, \( \triangle AOB \cong \triangle BOC \) (SAS axiom).
This means \( AB = BC \) (Corresponding parts of congruent triangles).
Since ABCD is a parallelogram, \( AB = CD \) and \( BC = AD \) (Opposite sides of parallelogram).
Because \( AB = BC \), it implies all sides are equal: \( AB = BC = CD = DA \). So, ABCD is a rhombus.
Now, in \( \triangle ABC \) and \( \triangle BAD \):
\( AB = AB \) (Common side).
\( AC = BD \) (Given, diagonals are equal).
\( BC = AD \) (Opposite sides of parallelogram).
So, \( \triangle ABC \cong \triangle BAD \) (SSS axiom).
This means \( \angle ABC = \angle BAD \) (Corresponding parts of congruent triangles).
Since ABCD is a parallelogram, \( \angle ABC + \angle BAD = 180^\circ \) (Consecutive interior angles).
As \( \angle ABC = \angle BAD \), we have \( 2\angle ABC = 180^\circ \), so \( \angle ABC = 90^\circ \).
A rhombus with one angle 90 degrees is a square. Hence, ABCD is a square.

(ii) In a rectangle, the diagonals are of equal lengths.

Given: ABCD is a rectangle. AC and BD are its diagonals.
To prove: \( AC = BD \).
Proof:
In \( \triangle ABC \) and \( \triangle BAD \):
\( AB = AB \) (Common side).
\( BC = AD \) (Opposite sides of a rectangle are equal).
\( \angle ABC = \angle BAD = 90^\circ \) (Angles of a rectangle are 90 degrees).
So, \( \triangle ABC \cong \triangle BAD \) (SAS axiom).
Therefore, \( AC = BD \) (Corresponding parts of congruent triangles). The diagonals of a rectangle are always equal in length.

(iii) In the figure, BM bisects \( \angle B \) and AN bisects \( \angle A \) of parallelogram ABCD. Prove that ACXM is an isosceles triangle.

Given: ABCD is a parallelogram. ABL and ADXY are squares.
To prove: ACXM is an isosceles triangle (implied to be \( \triangle CMX \) where CM = CX).
Construction: Join MC, MX and CX.
Proof:
In parallelogram ABCD, we know that:
\( AB = CD \) (Opposite sides)
\( BC = AD \) (Opposite sides)
\( \angle ABC = \angle CDA \) (Opposite angles)
Since ABL is a square, \( \angle LAB = 90^\circ \) (if ABL were ABML). Given ABL, it is interpreted as ABML.
If ABL is a square, then \( BM = AB \). Also, \( \angle ABM = 90^\circ \).
So, \( \angle MBC = \angle ABC + \angle ABM = \angle ABC + 90^\circ \).
Since ADXY is a square, \( DX = AD \). Also, \( \angle ADX = 90^\circ \).
So, \( \angle CDX = \angle CDA + \angle ADX = \angle CDA + 90^\circ \).
Since \( \angle ABC = \angle CDA \) (opposite angles of a parallelogram), it follows that \( \angle MBC = \angle CDX \).
Now consider \( \triangle BMC \) and \( \triangle DXC \):
\( BM = AB \) (Sides of square ABL). We know \( AB = CD \) (Opposite sides of parallelogram).
So, \( BM = CD \).
\( BC = AD \) (Opposite sides of parallelogram). We know \( AD = DX \) (Sides of square ADXY).
So, \( BC = DX \).
We have \( BM = CD \), \( BC = DX \), and \( \angle MBC = \angle CDX \) (proved above).
Therefore, \( \triangle BMC \cong \triangle DXC \) (SAS axiom).
By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), \( CM = CX \).
Since two sides of \( \triangle CMX \) are equal (\( CM = CX \)), \( \triangle CMX \) is an isosceles triangle.
Thus, the quadrilateral ACXM forms an isosceles triangle (CMX is isosceles).

(iv) In the figure, ABCD and ALMN are two squares. Prove that BL = DN.

Given: ABCD and ALMN are two squares.
To prove: \( BL = DN \).
Proof:
Consider \( \triangle ADN \) and \( \triangle ALB \):
\( AD = AB \) (Sides of square ABCD).
\( AN = AL \) (Sides of square ALMN).
Now for the included angles:
\( \angle DAN = \angle DAB + \angle BAN \). Since ABCD is a square, \( \angle DAB = 90^\circ \). So \( \angle DAN = 90^\circ + \angle BAN \).
\( \angle LAB = \angle LAN + \angle NAB \). Since ALMN is a square, \( \angle LAN = 90^\circ \). So \( \angle LAB = 90^\circ + \angle NAB \).
From these, we can see that \( \angle DAN = \angle LAB \). This means the included angles are equal.
Therefore, \( \triangle ADN \cong \triangle ALB \) (SAS axiom).
By CPCTC, \( DN = BL \). Both BL and DN are diagonals of congruent triangles, connecting a vertex of one square to a vertex of the other.
In simple words: If you have two squares that share one corner (point A), and you draw a line from one corner of the first square to the opposite corner of the second square (BL), and another line from the other corner of the first square to the opposite corner of the second square (DN), these two lines will be the same length. This is because the triangles they form are identical.

🎯 Exam Tip: When dealing with overlapping squares, always look for congruent triangles using SAS, SSS, or ASA criteria, especially by identifying equal sides from the square properties and common or sum of angles.

Question 12.
(i) Prove that the bisectors of any two adjacent angles of a parallelogram are at right angles.
(ii) In the figure, PQRS is a parallelogram with bisectors PA, QD, RC and SB respectively of angles P, Q, R and S. Show that ABCD is an rectangle.

Answer:
(i) Given: ABCD is a parallelogram. AE is the bisector of \( \angle A \) and BE is the bisector of \( \angle B \). Both bisectors meet at E.

To prove: \( \angle AEB = 90^\circ \).
Proof:
In parallelogram ABCD, adjacent angles are supplementary.
So, \( \angle A + \angle B = 180^\circ \).
AE bisects \( \angle A \), which means \( \angle EAB = \frac{1}{2} \angle A \).
BE bisects \( \angle B \), which means \( \angle EBA = \frac{1}{2} \angle B \).
Now, consider \( \triangle AEB \). The sum of its angles is 180 degrees.
\( \angle AEB + \angle EAB + \angle EBA = 180^\circ \).
Substitute the bisected angles:
\( \angle AEB + \frac{1}{2} \angle A + \frac{1}{2} \angle B = 180^\circ \).
\( \angle AEB + \frac{1}{2} (\angle A + \angle B) = 180^\circ \).
Since \( \angle A + \angle B = 180^\circ \), we have:
\( \angle AEB + \frac{1}{2} (180^\circ) = 180^\circ \).
\( \angle AEB + 90^\circ = 180^\circ \).
\( \angle AEB = 180^\circ - 90^\circ = 90^\circ \).
Thus, the bisectors of any two adjacent angles of a parallelogram meet at a right angle.

(ii) Given: PQRS is a parallelogram. PA, QD, RC, and SB are the bisectors of \( \angle P, \angle Q, \angle R \), and \( \angle S \) respectively, forming a quadrilateral ABCD where they meet. To prove: ABCD is a rectangle.
Proof:
In parallelogram PQRS, adjacent angles are supplementary.
Consider \( \angle S \) and \( \angle P \). Their bisectors SB and PA meet at A.
Since \( \angle S + \angle P = 180^\circ \), from part (i) we know that the angle formed by their bisectors, \( \angle SAD + \angle PSA \) (or \( \angle A \)), is 90 degrees. Thus, \( \angle A = 90^\circ \).
Similarly, for \( \angle P \) and \( \angle Q \). Their bisectors PA and QD meet at D.
Since \( \angle P + \angle Q = 180^\circ \), the angle formed by their bisectors, \( \angle APQ + \angle DQP \) (or \( \angle D \)), is 90 degrees. Thus, \( \angle D = 90^\circ \).
Similarly, for \( \angle Q \) and \( \angle R \). Their bisectors QD and RC meet at C.
Since \( \angle Q + \angle R = 180^\circ \), the angle formed by their bisectors, \( \angle RCD + \angle CQD \) (or \( \angle C \)), is 90 degrees. Thus, \( \angle C = 90^\circ \).
Similarly, for \( \angle R \) and \( \angle S \). Their bisectors RC and SB meet at B.
Since \( \angle R + \angle S = 180^\circ \), the angle formed by their bisectors, \( \angle RBS + \angle SCR \) (or \( \angle B \)), is 90 degrees. Thus, \( \angle B = 90^\circ \).
Since all four angles of quadrilateral ABCD are 90 degrees, ABCD is a rectangle.
In simple words: When you take any parallelogram and draw lines that cut each of its inside angles exactly in half, these four new lines will cross each other to form a perfect rectangle in the middle.

🎯 Exam Tip: This is a standard proof. Remember that adjacent angles of a parallelogram add up to 180°, and half of that sum is 90°, which is the angle formed by their bisectors.

Question 13. In the figure, ABCD is a parallelogram and X is the midpoint of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that:
(i) \( \triangle ABX \cong \triangle QCX \)

Answer:
(i) Given: ABCD is a parallelogram. X is the mid-point of BC. The line AX is produced to meet DC produced at Q. Parallelogram ABPQ is completed.
To prove: \( \triangle ABX \cong \triangle QCX \).
Proof:
Consider \( \triangle ABX \) and \( \triangle QCX \):
\( BX = XC \) (Given that X is the mid-point of BC).
\( \angle AXB = \angle QXC \) (These are vertically opposite angles, which are always equal).
Since ABCD is a parallelogram, AB is parallel to DC. This means AB is also parallel to DQ (DC produced).
Now, AXQ is a transversal cutting parallel lines AB and DQ.
Therefore, \( \angle BAX = \angle CQX \) (These are alternate interior angles).
By the AAS (Angle-Angle-Side) axiom of congruence, we can say that:
\( \triangle ABX \cong \triangle QCX \).
From this congruence, we also get \( AB = QC \) (Corresponding parts of congruent triangles).
Also, in parallelogram ABCD, \( AB = DC \) (Opposite sides).
And in parallelogram ABPQ, \( AB = QP \) (Opposite sides).
Therefore, \( DC = CQ = QP \), which means D, C, Q are collinear and C is the midpoint of DQ.
In simple words: Imagine a tilted box (parallelogram) and a line drawn from one corner (A) to the middle of the opposite side (X). If you keep drawing that line straight past X until it hits the extended top side of the box (Q), it creates two identical triangles (\( \triangle ABX \) and \( \triangle QCX \)).

🎯 Exam Tip: For congruence proofs, always clearly state the three matching conditions (side/angle) and the correct congruence axiom (SAS, ASA, AAS, SSS, RHS) you are using. Identify vertically opposite and alternate interior angles carefully.

Question 14. ABCD is a rhombus with P, Q, R as mid-points of AB, BC and CD. Prove that PQ \( \perp \) QR.

Answer:
Given: ABCD is a rhombus. P, Q, and R are the mid-points of sides AB, BC, and CD respectively. PQ and QR are joined.
To prove: PQ \( \perp \) QR.
Construction: Join the diagonals AC and BD of the rhombus ABCD.
Proof:
In \( \triangle ABC \), P is the mid-point of AB and Q is the mid-point of BC.
By the Mid-point Theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of its length.
So, \( PQ \parallel AC \) and \( PQ = \frac{1}{2} AC \). (Equation i)
In \( \triangle BCD \), Q is the mid-point of BC and R is the mid-point of CD.
Similarly, by the Mid-point Theorem:
So, \( QR \parallel BD \) and \( QR = \frac{1}{2} BD \). (Equation ii)
We know that in a rhombus, the diagonals intersect each other at right angles.
Therefore, \( AC \perp BD \).
Since \( PQ \parallel AC \) (from (i)) and \( QR \parallel BD \) (from (ii)), and \( AC \perp BD \), it follows that PQ must be perpendicular to QR.
When two lines are parallel to two other lines that are perpendicular, the first two lines are also perpendicular to each other.
Thus, \( PQ \perp QR \).
In simple words: In a diamond shape (rhombus), if you connect the middle points of three sides in a row, the lines you draw will meet at a perfect right angle. This happens because the diagonals of a rhombus always cross at 90 degrees, and our new lines are parallel to these diagonals.

🎯 Exam Tip: The Mid-point Theorem is crucial here. Remember that if line1 \( \parallel \) line3 and line2 \( \parallel \) line4, then if line3 \( \perp \) line4, it implies line1 \( \perp \) line2.

Question 15. ABCD is a rhombus. RABS is a straight line such that RA = AB = BS. Prove that RD and SC when produced meet at right angles.

Answer:
Given: ABCD is a rhombus. RABS is a straight line such that \( RA = AB = BS \). RD and SC are produced to meet at T.
To prove: \( \angle T = 90^\circ \).
Proof:
In a rhombus ABCD, all sides are equal: \( AB = BC = CD = DA \).
Given that \( RA = AB = BS \).
From these two facts, we have: \( RA = AD \) and \( BS = BC \).
Consider \( \triangle RAD \):
Since \( RA = AD \), \( \triangle RAD \) is an isosceles triangle.
Therefore, \( \angle RDA = \angle ARD \).
The exterior angle \( \angle DAB \) of \( \triangle RAD \) is equal to the sum of the two opposite interior angles.
\( \angle DAB = \angle RDA + \angle ARD = 2\angle RDA \).
So, \( \angle RDA = \frac{1}{2} \angle DAB \). (Let \( \angle DAB = 2\alpha \), then \( \angle RDA = \alpha \)).
Consider \( \triangle BSC \):
Since \( BS = BC \), \( \triangle BSC \) is an isosceles triangle.
Therefore, \( \angle BSC = \angle BCS \).
The exterior angle \( \angle ABC \) of \( \triangle BSC \) is equal to the sum of the two opposite interior angles.
\( \angle ABC = \angle BSC + \angle BCS = 2\angle BSC \).
So, \( \angle BSC = \frac{1}{2} \angle ABC \). (Let \( \angle ABC = 2\beta \), then \( \angle BSC = \beta \)).
In rhombus ABCD, adjacent angles are supplementary:
\( \angle DAB + \angle ABC = 180^\circ \).
Substituting \( 2\alpha \) and \( 2\beta \): \( 2\alpha + 2\beta = 180^\circ \).
Dividing by 2, we get: \( \alpha + \beta = 90^\circ \).
Now, consider \( \triangle RST \) (where R and S are points on the line RABS, and T is the intersection of RD and SC produced):
The angle \( \angle TRS \) is the same as \( \angle ARD \) from \( \triangle RAD \). So, \( \angle TRS = \alpha \).
The angle \( \angle TSR \) is the same as \( \angle BSC \) from \( \triangle BSC \). So, \( \angle TSR = \beta \).
The sum of angles in \( \triangle RST \) is 180 degrees:
\( \angle RTS + \angle TRS + \angle TSR = 180^\circ \).
\( \angle RTS + \alpha + \beta = 180^\circ \).
Substitute \( \alpha + \beta = 90^\circ \):
\( \angle RTS + 90^\circ = 180^\circ \).
\( \angle RTS = 180^\circ - 90^\circ = 90^\circ \).
Therefore, RD and SC when produced meet at right angles.
In simple words: Imagine a diamond shape (rhombus). If you extend its base line evenly on both sides, then draw lines from the top corners of the rhombus to the ends of this extended base, these two new lines will cross each other at a perfect right angle. This happens because the angles created at the base add up to 90 degrees.

🎯 Exam Tip: This question involves using properties of rhombuses (equal sides, supplementary adjacent angles) and isosceles triangles (equal base angles). The key is to relate the interior and exterior angles correctly to find the sum of angles in the final triangle.