OP Malhotra Class 9 Maths Solutions Chapter 9 Mid Point and Intercept Theorems Exercise 9 (A)

Question 1. Fill in the blanks :
(i) The line joining the mid-points of two sides of a triangle is .................... to the third side.
(ii) The line drawn through the mid-point of one side of a triangle parallel to another side bisects the ................ side.
(iii) In figure, S and T are the mid-points of PQ and PR respectively. If ST = 3 cm, then QR =
(iv) In figure, D and E are the mid-points of AB and AC. If DE = 7.5 cm, then BC = ....................
(v) Find the value of x in each triangle.
Answer:
(i) The line joining the mid-points of two sides of a triangle is **parallel** to the third side.
(ii) The line drawn through the mid-point of one side of a triangle parallel to another side bisects the **third** side.
(iii) In the figure, S and T are the mid-points of PQ and PR respectively. If ST = 3 cm, then QR = **6 cm**. This is because the line segment connecting the mid-points of two sides of a triangle is half the length of the third side.
(iv) In the figure, D and E are the mid-points of AB and AC. If DE = 7.5 cm, then BC = **15 cm**. The segment connecting midpoints is half the length of the third side, so the third side is twice that length.
(v) (a) In the given triangle, from the midpoint theorem, if D and E are midpoints of AB and BC respectively, then DE is parallel to AC and \( DE = \frac{1}{2} AC \).
Here, \( DE = 64 \) and \( AC = 4x \).
So, \( 64 = \frac{1}{2} \times 4x \)
\( 64 = 2x \)
\( \implies x = \frac{64}{2} \)
\( \implies x = 32 \)
(b) In the given triangle, if D and E are midpoints of AB and AC respectively, then DE is parallel to BC and \( DE = \frac{1}{2} BC \).
Here, \( DE = x-8 \) and \( BC = 40 \).
So, \( x-8 = \frac{1}{2} \times 40 \)
\( x-8 = 20 \)
\( \implies x = 20 + 8 \)
\( \implies x = 28 \)
(c) In the given triangle, if D and E are midpoints of AC and BC respectively, then DE is parallel to AB and \( DE = \frac{1}{2} AB \).
Here, \( DE = x+8 \) and \( AB = 6x \).
So, \( x+8 = \frac{1}{2} \times 6x \)
\( x+8 = 3x \)
\( \implies 3x - x = 8 \)
\( 2x = 8 \)
\( \implies x = \frac{8}{2} \)
\( \implies x = 4 \)
In simple words: The Midpoint Theorem tells us that a line connecting the middle points of two sides of a triangle is always parallel to the third side and exactly half its length. We use this rule to fill in the blanks and find unknown lengths in the triangles.

🎯 Exam Tip: Remember the two key parts of the Midpoint Theorem: the line segment is parallel to the third side, and its length is exactly half that of the third side. Both aspects are important for proofs and calculations.

Question 2. In ∆ABC, AB = 6 cm, and AC = 3 cm. If M is the mid-point of AB, and a straight line through M parallel to AC cuts BC in N, what is the length of MN?
Answer: Given ∆ABC, AB = 6 cm, and AC = 3 cm. M is the mid-point of AB, and a line through M parallel to AC cuts BC at N.
Since M is the mid-point of AB and MN is parallel to AC, by the converse of the Midpoint Theorem, N must be the mid-point of BC.
Also, by the Midpoint Theorem, the length of MN will be half the length of AC.
So, \( MN = \frac{1}{2} \times AC \)
\( MN = \frac{1}{2} \times 3 \text{ cm} \)
\( MN = 1.5 \text{ cm} \)
In simple words: In this triangle, M is in the middle of side AB, and a line from M runs parallel to side AC and touches BC at N. Because of a special rule (the converse of the Midpoint Theorem), N must also be in the middle of BC. This means the line MN will be half as long as AC.

🎯 Exam Tip: When a line starts from the midpoint of one side and runs parallel to another side, remember it will always bisect the third side. This is the converse of the Midpoint Theorem and is often used in problems.

Question 2. In quadrilateral ∆ABC, D is the mid-point of AB, E is the mid-point of AC. Calculate:
(i) DE, if BC = 6 cm
(ii) ∠ADE, if ∠DBC = 140°
Answer: In the given figure, ∆ABC has D as the midpoint of AB and E as the midpoint of AC.
(i) According to the Midpoint Theorem, the line segment connecting the mid-points of two sides of a triangle is half the length of the third side and parallel to it.
Given BC = 6 cm.
Therefore, \( DE = \frac{1}{2} BC \)
\( DE = \frac{1}{2} \times 6 \text{ cm} \)
\( DE = 3 \text{ cm} \)
(ii) Given ∠DBC = 140°.
Since DE is parallel to BC (from the Midpoint Theorem), and DB is a transversal line, the angles ∠ADE and ∠DBC are corresponding angles. Corresponding angles are equal when two parallel lines are cut by a transversal.
So, \( \angle ADE = \angle DBC \)
\( \angle ADE = 140^\circ \)
In simple words: First, DE is half of BC because D and E are midpoints. Second, since DE is parallel to BC, the angle ∠ADE is the same as ∠DBC because they are matching angles formed by a line cutting two parallel lines.

🎯 Exam Tip: Always remember that corresponding angles are equal when parallel lines are involved. Clearly stating the parallel lines and the transversal helps in justifying angle equalities.

Question 4. In figure, D, E, F are the mid-points of BC, CA and AB respectively. Prove that AD bisects EF.
Answer: Given that D, E, and F are the mid-points of the sides BC, CA, and AB respectively.
To prove that AD bisects EF, we first join ED and FD.
In ∆ABC,
Since E and F are the mid-points of CA and AB, by the Midpoint Theorem, EF is parallel to BC and \( EF = \frac{1}{2} BC \).
Similarly,
Since E and D are the mid-points of AC and BC, ED is parallel to AB and \( ED = \frac{1}{2} AB \).
Since F and D are the mid-points of AB and BC, FD is parallel to AC and \( FD = \frac{1}{2} AC \).
Because EF || BC, ED || AB, and FD || AC, the quadrilateral BDEF is a parallelogram (since opposite sides are parallel). Also, AEDF forms a parallelogram due to these properties. Consider the quadrilateral BDEF, where EF || BD (part of BC) and ED || BF (part of AB). This means BDEF is a parallelogram.
The diagonals of a parallelogram bisect each other.
Therefore, AD (which is part of the diagonal of BDEF or ADEF) must bisect EF.
In simple words: When you connect the middle points of a triangle's sides, you create smaller parallelograms. Since AD is a diagonal of one of these parallelograms (like AEDF), it must cut the other diagonal (EF) exactly in half.

🎯 Exam Tip: To prove that one line bisects another, a common strategy is to show that a parallelogram is formed, as the diagonals of a parallelogram always bisect each other. Identify the parallelogram first.

Question 5. In figure, ABC is an isosceles triangle in which AB = AC. D, E and F are respectively the mid-points of sides BC, AB and CA. Show that the line segments AD, EF bisect each other at right angles.
Answer: Given that ABC is an isosceles triangle where AB = AC. D, E, and F are the mid-points of BC, AB, and CA respectively.
We need to prove that AD and EF bisect each other at right angles.
From the given information, E and F are mid-points of AB and AC.
So, AE = EB and AF = FC.
Since AB = AC (given), it follows that \( \frac{1}{2}AB = \frac{1}{2}AC \), which means AE = AF.
By the Midpoint Theorem, EF is parallel to BC and \( EF = \frac{1}{2} BC \).
Now consider ∆ABD and ∆ACD:
AB = AC (given)
BD = DC (since D is the mid-point of BC)
AD = AD (common side)
Therefore, ∆ABD ≅ ∆ACD (by SSS axiom).
This implies ∠BAD = ∠CAD (by c.p.c.t.). So, AD bisects ∠BAC. Since AD is the median to the base of an isosceles triangle, it is also the altitude, meaning AD is perpendicular to BC. Because EF is parallel to BC, AD must also be perpendicular to EF.
Now consider ∆AEG and ∆AFG, where G is the intersection of AD and EF.
AE = AF (as shown earlier, \( \frac{1}{2} \) of equal sides)
∠EAG = ∠FAG (since AD bisects ∠BAC)
AG = AG (common side)
Therefore, ∆AEG ≅ ∆AFG (by SAS axiom).
This implies ∠AGE = ∠AGF (by c.p.c.t.).
Since ∠AGE and ∠AGF form a linear pair, their sum is 180°.
\( \angle AGE + \angle AGF = 180^\circ \)
Since they are equal, \( \angle AGE = \angle AGF = 90^\circ \).
This means AD is perpendicular to EF.
Since E and F are midpoints, EF || BC. Also AD is the median to the base BC in isosceles ∆ABC, so AD is perpendicular to BC. Because EF is parallel to BC, AD is also perpendicular to EF. In parallelogram ADEF (as proved in Q4), diagonals AD and EF bisect each other. So, we have shown that AD and EF bisect each other at right angles.
In simple words: Since the triangle ABC is isosceles (AB=AC), the line AD from the top corner to the middle of the base is also straight up and down (perpendicular) to the base. The line EF connects the middle points of the other two sides, making it parallel to the base. Because AD is perpendicular to the base, it's also perpendicular to EF. Also, in the quadrilateral AEFD, the lines AD and EF act as diagonals and cut each other in half.

🎯 Exam Tip: For isosceles triangles, remember that the median to the base is also the altitude and the angle bisector. This property, combined with the Midpoint Theorem, is crucial for proving perpendicularity and bisection.

Question 6. Prove that the four triangles formed by joining the pairs of mid-points of the three sides of a triangle are congruent to each other.
Answer: Given ∆ABC, D, E, and F are the mid-points of the sides BC, CA, and AB respectively. DE, EF, and FD are joined.
We need to prove that ∆AEF ≅ ∆BDF ≅ ∆CDE ≅ ∆DEF.
Proof:
1. Consider ∆AEF:
E and F are the mid-points of AC and AB respectively.
By the Midpoint Theorem, EF || BC and \( EF = \frac{1}{2} BC \).
Similarly, F and D are the mid-points of AB and BC respectively.
So, FD || AC and \( FD = \frac{1}{2} AC \).
And D and E are the mid-points of BC and CA respectively.
So, DE || AB and \( DE = \frac{1}{2} AB \).
Now, in ∆AEF and ∆DEF:
We know that \( AE = \frac{1}{2} AB \). From above, \( DE = \frac{1}{2} AB \). So, AE = DE.
We know that \( AF = \frac{1}{2} AB \). From above, \( FD = \frac{1}{2} AC \). This part of the proof has a slight logical jump in the source. Let's re-establish properly.
From Midpoint Theorem:
\( DE = \frac{1}{2} AB \) (D and E are mid-points of BC and AC)
\( EF = \frac{1}{2} BC \) (E and F are mid-points of AC and AB)
\( FD = \frac{1}{2} AC \) (F and D are mid-points of AB and BC)

Now compare the triangles:
In ∆AEF:
\( AF = \frac{1}{2} AB \)
\( AE = \frac{1}{2} AC \)
\( EF = \frac{1}{2} BC \)

In ∆BDF:
\( BF = \frac{1}{2} AB \)
\( BD = \frac{1}{2} BC \)
\( FD = \frac{1}{2} AC \)

In ∆CDE:
\( CD = \frac{1}{2} BC \)
\( CE = \frac{1}{2} AC \)
\( DE = \frac{1}{2} AB \)

In ∆DEF:
\( DE = \frac{1}{2} AB \)
\( EF = \frac{1}{2} BC \)
\( FD = \frac{1}{2} AC \)

By comparing the sides:
1. \( AF = \frac{1}{2} AB \), and \( DE = \frac{1}{2} AB \). So, AF = DE.
2. \( AE = \frac{1}{2} AC \), and \( FD = \frac{1}{2} AC \). So, AE = FD.
3. \( EF = \frac{1}{2} BC \), and \( BD = \frac{1}{2} BC \) (since D is midpoint). So, EF = BD.
Similarly, \( EF = \frac{1}{2} BC \), and \( CD = \frac{1}{2} BC \). So, EF = CD.

Consider ∆AEF and ∆DEF:
AE = FD (Proved above)
AF = DE (Proved above)
EF = EF (Common side)
Therefore, ∆AEF ≅ ∆DEF (by SSS criterion).

Consider ∆BDF and ∆DEF:
BF = DE (Since F is midpoint of AB, \( BF = \frac{1}{2} AB \). And \( DE = \frac{1}{2} AB \). So, BF = DE)
BD = EF (Since D is midpoint of BC, \( BD = \frac{1}{2} BC \). And \( EF = \frac{1}{2} BC \). So, BD = EF)
FD = FD (Common side)
Therefore, ∆BDF ≅ ∆DEF (by SSS criterion).

Consider ∆CDE and ∆DEF:
CD = EF (Since D is midpoint of BC, \( CD = \frac{1}{2} BC \). And \( EF = \frac{1}{2} BC \). So, CD = EF)
CE = FD (Since E is midpoint of AC, \( CE = \frac{1}{2} AC \). And \( FD = \frac{1}{2} AC \). So, CE = FD)
DE = DE (Common side)
Therefore, ∆CDE ≅ ∆DEF (by SSS criterion).

Since ∆AEF ≅ ∆DEF, ∆BDF ≅ ∆DEF, and ∆CDE ≅ ∆DEF, it means all four triangles (∆AEF, ∆BDF, ∆CDE, and ∆DEF) are congruent to each other.
In simple words: When you connect the middle points of all three sides of any triangle, it divides the big triangle into four smaller triangles. Each of these four small triangles will have sides that are exactly half the length of the original triangle's sides, just in a different order. Because all their corresponding sides are equal, all four small triangles are exactly the same shape and size (congruent).

🎯 Exam Tip: This problem uses the Midpoint Theorem repeatedly. For full marks, clearly state which segments are parallel and half the length of the third side based on the midpoints, and then use the SSS congruence rule to show the triangles are congruent.

Question 7. D, E and F are respectively the mid-points of the sides BC, CA and AB of an equilateral triangle. Prove that ∆DEF is also an equilateral triangle.
Answer: Given ∆ABC is an equilateral triangle. D, E, and F are the mid-points of sides BC, CA, and AB respectively. DE, EF, and FD are joined.
We need to prove that ∆DEF is an equilateral triangle.
Proof:
Since E and F are the mid-points of AC and AB respectively, by the Midpoint Theorem:
EF || BC and \( EF = \frac{1}{2} BC \) ... (i)
Similarly, since D and E are the mid-points of BC and AC respectively, by the Midpoint Theorem:
DE || AB and \( DE = \frac{1}{2} AB \) ... (ii)
And since D and F are the mid-points of BC and AB respectively, by the Midpoint Theorem:
FD || AC and \( FD = \frac{1}{2} AC \) ... (iii)
We know that ∆ABC is an equilateral triangle, so all its sides are equal:
AB = BC = CA
From (i), (ii), and (iii):
\( EF = \frac{1}{2} BC \)
\( DE = \frac{1}{2} AB \)
\( FD = \frac{1}{2} AC \)
Since AB = BC = CA, it follows that \( \frac{1}{2} AB = \frac{1}{2} BC = \frac{1}{2} CA \).
Therefore, EF = DE = FD.
Since all three sides of ∆DEF are equal, ∆DEF is an equilateral triangle.
In simple words: If you start with a triangle where all sides are the same length (an equilateral triangle) and then connect the middle points of each side, you will form a new, smaller triangle in the middle. Because of the Midpoint Theorem, each side of this new small triangle will be exactly half the length of the original triangle's sides. Since the original sides were all equal, the new sides will also all be equal, making the small inner triangle an equilateral triangle too.

🎯 Exam Tip: The key to this proof is realizing that if the original triangle's sides are equal, then half of those equal sides will also be equal. This directly proves the equilateral nature of the inner triangle formed by midpoints.

Question 8. In figure, AD and BE are the medians of ∆ABC and BE || DF. Prove that \( CF = \frac{1}{4} AC \).
Answer: Given that AD and BE are medians of ∆ABC, and BE || DF.
To prove: \( CF = \frac{1}{4} AC \).
Proof:
Since BE is a median, E is the mid-point of AC.
Therefore, \( AE = EC \) or \( EC = \frac{1}{2} AC \) ... (i)
Now consider ∆BCE:
D is the mid-point of BC (since AD is a median).
It is given that BE || DF. This means that DF is parallel to BE.
Since D is the mid-point of BC and DF || BE, by the converse of the Midpoint Theorem (also known as the intercept theorem for triangles), F must be the mid-point of EC.
Therefore, \( FC = \frac{1}{2} EC \) ... (ii)
Now, substitute the value of EC from (i) into (ii):
\( CF = \frac{1}{2} \left( \frac{1}{2} AC \right) \)
\( CF = \frac{1}{4} AC \)
Hence proved.
In simple words: Since E is the middle of AC, EC is half of AC. Next, in the triangle BCE, D is the middle of BC. Because the line DF is parallel to BE, F must be the middle of EC. So, CF is half of EC. If EC is half of AC, then CF must be half of half of AC, which means CF is one-fourth of AC.

🎯 Exam Tip: This question elegantly combines the definition of a median with the converse of the Midpoint Theorem. Clearly state why E and D are midpoints, and then apply the theorem to establish F as a midpoint.

Question 9. In the given figure, ABC is an isosceles ∆ with AB = AC and CP || BA and AP is the bisector of ext. ∠CAD of ∆ABC. Prove that
(i) ∠PAC = ∠BCA
(ii) ABCP is a parallelogram.
Answer: Given ABC is an isosceles triangle with AB = AC. CP || BA is drawn, and AP is the bisector of the exterior angle ∠CAD.
(i) To prove: ∠PAC = ∠BCA
Proof:
In ∆ABC, since AB = AC, the angles opposite to these equal sides are equal.
So, ∠C = ∠B (or ∠BCA = ∠ABC).
The exterior angle ∠CAD is equal to the sum of the two interior opposite angles.
So, ∠CAD = ∠ABC + ∠BCA.
Since ∠ABC = ∠BCA, we can write ∠CAD = ∠BCA + ∠BCA = 2∠BCA ... (1)
It is given that AP bisects ∠CAD, which means it divides the angle into two equal halves.
So, \( \angle PAC = \angle PAD = \frac{1}{2} \angle CAD \) ... (2)
From (1) and (2):
\( \angle PAC = \frac{1}{2} (2 \angle BCA) \)
\( \angle PAC = \angle BCA \).
Thus, ∠PAC = ∠BCA is proved.
(ii) To prove: ABCP is a parallelogram.
Proof:
From part (i), we have proved that ∠PAC = ∠BCA.
These are alternate interior angles for lines AP and BC with transversal AC. Since alternate interior angles are equal, this implies AP || BC.
It is also given that CP || BA.
Since both pairs of opposite sides are parallel (AP || BC and CP || BA), the quadrilateral ABCP is a parallelogram.
Hence proved.
In simple words: First, because triangle ABC has two equal sides (AB=AC), the angles opposite these sides are also equal (∠B=∠C). The angle outside the triangle (∠CAD) is equal to the sum of the two opposite inside angles (∠B + ∠C), which simplifies to 2∠C. Since AP cuts this outside angle exactly in half, ∠PAC is half of ∠CAD, meaning ∠PAC = ∠C. This proves the first part. For the second part, since ∠PAC = ∠BCA (which are alternate angles), it means AP is parallel to BC. We are also given that CP is parallel to BA. Because both pairs of opposite sides are parallel, the shape ABCP is a parallelogram.

🎯 Exam Tip: When proving a quadrilateral is a parallelogram, showing that both pairs of opposite sides are parallel is a direct method. Use properties of isosceles triangles (angles opposite equal sides are equal) and exterior angles of a triangle to establish angle equalities, then use alternate interior angles to prove parallel lines.

Question 10. ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the midpoints of the sides, in order, is a rectangle.
Answer: Given ABCD is a kite where AB = AD and BC = CD. Let P, Q, R, and S be the mid-points of the sides AB, BC, CD, and DA respectively. PQ, QR, RS, and SP are joined.
We need to prove that PQRS is a rectangle.
Construction: Join BD and AC (the diagonals of the kite).
Proof:
1. In ∆ABD:
P and Q are the mid-points of AB and AD respectively.
By the Midpoint Theorem, PQ || BD and \( PQ = \frac{1}{2} BD \) ... (i)
2. In ∆BCD:
S and R are the mid-points of BC and CD respectively.
By the Midpoint Theorem, SR || BD and \( SR = \frac{1}{2} BD \) ... (ii)
From (i) and (ii), we can conclude that PQ || SR and PQ = SR. This means one pair of opposite sides of PQRS is both parallel and equal, so PQRS is a parallelogram.
3. Similarly, in ∆ABC, P and S are mid-points of AB and BC respectively.
So, PS || AC and \( PS = \frac{1}{2} AC \).
4. In ∆ADC, Q and R are mid-points of AD and CD respectively.
So, QR || AC and \( QR = \frac{1}{2} AC \).
Therefore, PS || QR and PS = QR. This further confirms PQRS is a parallelogram.
5. Properties of a kite: The diagonals of a kite intersect each other at right angles.
So, AC \( \perp \) BD.
We know that PQ || BD and PS || AC.
Since AC \( \perp \) BD, and PS is parallel to AC, and PQ is parallel to BD, it implies that PS \( \perp \) PQ (because if two lines are perpendicular, any lines parallel to them will also be perpendicular).
Therefore, ∠SPQ = 90°.
Since PQRS is a parallelogram with one angle equal to 90°, it must be a rectangle.
Hence proved.
In simple words: A kite has two pairs of equal sides that are next to each other. When you connect the middle points of all its sides, you get a new shape called PQRS. Using the Midpoint Theorem, we can show that opposite sides of PQRS are parallel and equal, meaning it's a parallelogram. Also, the special property of a kite is that its main diagonals cross each other at a right angle. Since the sides of PQRS are parallel to these diagonals, the sides of PQRS will also meet at a right angle. A parallelogram with a right angle is a rectangle.

🎯 Exam Tip: To prove a quadrilateral formed by midpoints is a rectangle, first prove it's a parallelogram using the Midpoint Theorem. Then, use the property of the diagonals of the original figure (a kite, in this case, having perpendicular diagonals) to show that one of the angles of the parallelogram is 90 degrees.

Question 11. Show that the quadrilateral, formed by joining the mid-points of the consecutive sides of a rectangle, is a rhombus.
Answer: Given ABCD is a rectangle. P, Q, R, and S are the mid-points of the sides AB, BC, CD, and DA respectively. PQ, QR, RS, and SP are joined.
We need to prove that PQRS is a rhombus.
Construction: Join AC and BD (the diagonals of the rectangle).
Proof:
1. In ∆ABC:
P and Q are the mid-points of AB and BC respectively.
By the Midpoint Theorem, PQ || AC and \( PQ = \frac{1}{2} AC \) ... (i)
2. In ∆ADC:
S and R are the mid-points of AD and CD respectively.
By the Midpoint Theorem, SR || AC and \( SR = \frac{1}{2} AC \) ... (ii)
From (i) and (ii), we get PQ || SR and PQ = SR.
This means that PQRS is a parallelogram.
3. Similarly, consider the other pair of sides:
In ∆ABD, P and S are the mid-points of AB and AD respectively.
So, PS || BD and \( PS = \frac{1}{2} BD \).
In ∆BCD, Q and R are the mid-points of BC and CD respectively.
So, QR || BD and \( QR = \frac{1}{2} BD \).
Thus, PS || QR and PS = QR.
4. Properties of a rectangle: The diagonals of a rectangle are equal in length.
So, AC = BD.
From our earlier findings:
\( PQ = \frac{1}{2} AC \)
\( PS = \frac{1}{2} BD \)
Since AC = BD, it follows that \( PQ = PS \).
Since PQRS is a parallelogram and two adjacent sides (PQ and PS) are equal, it must be a rhombus.
Hence proved.
In simple words: If you take a rectangle and connect the middle points of all its sides, the new shape you get will be a rhombus. First, by connecting midpoints, we form a parallelogram. Then, since the diagonals of a rectangle are equal, and the sides of our new shape are half the length of these diagonals, all the sides of the new shape will be equal. A parallelogram with all equal sides is a rhombus.

🎯 Exam Tip: To show that the figure formed by joining midpoints is a rhombus, first establish that it is a parallelogram. Then, use the specific property of the rectangle's diagonals (they are equal) to prove that adjacent sides of the parallelogram are equal, thus making it a rhombus.

Question 12. Quad. ABCD is a rhombus and P, Q, R, S are the mid-points of AB, BC, CD, DA respectively. Prove that quad. PQRS is a rectangle.
Answer: Given ABCD is a rhombus. P, Q, R, and S are the mid-points of the sides AB, BC, CD, and DA respectively. PQ, QR, RS, and SP are joined.
We need to prove that PQRS is a rectangle.
Construction: Join AC and BD (the diagonals of the rhombus).
Proof:
1. In ∆ABC:
P and Q are the mid-points of AB and BC respectively.
By the Midpoint Theorem, PQ || AC and \( PQ = \frac{1}{2} AC \) ... (i)
2. In ∆ADC:
S and R are the mid-points of AD and CD respectively.
By the Midpoint Theorem, SR || AC and \( SR = \frac{1}{2} AC \) ... (ii)
From (i) and (ii), we get PQ || SR and PQ = SR. This means PQRS is a parallelogram.
3. Similarly, consider the other pair of sides:
In ∆ABD, P and S are the mid-points of AB and AD respectively.
So, PS || BD and \( PS = \frac{1}{2} BD \).
In ∆BCD, Q and R are the mid-points of BC and CD respectively.
So, QR || BD and \( QR = \frac{1}{2} BD \).
Thus, PS || QR and PS = QR. This confirms PQRS is a parallelogram.
4. Properties of a rhombus: The diagonals of a rhombus bisect each other at right angles.
So, AC \( \perp \) BD.
We know that PQ || AC and PS || BD.
Since AC \( \perp \) BD, and PS is parallel to BD, and PQ is parallel to AC, it implies that PS \( \perp \) PQ (because if two lines are perpendicular, any lines parallel to them will also be perpendicular).
Therefore, ∠SPQ = 90°.
Since PQRS is a parallelogram with one angle equal to 90°, it must be a rectangle.
Hence proved.
In simple words: When you connect the middle points of all sides of a rhombus, the new shape you get is a rectangle. First, we show that this new shape is a parallelogram. Then, we use the fact that the diagonals of a rhombus always cross at a perfect right angle. Since the sides of our new shape are parallel to these diagonals, those sides will also meet at a right angle. A parallelogram with a right angle is a rectangle.

🎯 Exam Tip: The crucial step here is to use the property that the diagonals of a rhombus are perpendicular. This relationship, combined with the parallelism derived from the Midpoint Theorem, directly leads to proving a right angle in the inner quadrilateral.

Question 13. In the quadrilateral ABCD, the mid-points of AB, BC, CD, DA are L, M, P, Q.
(i) Using the mid-point theorem make a statement concerning the lengths and directions of LM and AC.
(ii) Prove that LMPQ is a parallelogram.
(iii) If it is also given that the diagonals AC and BD are equal. What further statement can be made about the parallelogram LMPQ ?
Answer: Given a quadrilateral ABCD, L, M, P, and Q are the mid-points of the sides AB, BC, CD, and DA respectively. LM and AC are joined.
(i) Using the Midpoint Theorem, a statement concerning the lengths and directions of LM and AC is:
In ∆ABC, L and M are the mid-points of AB and BC respectively.
Therefore, LM is parallel to AC (LM || AC), and the length of LM is half the length of AC (\( LM = \frac{1}{2} AC \)).
(ii) To prove that LMPQ is a parallelogram.
Proof:
From part (i), we know LM || AC and \( LM = \frac{1}{2} AC \) ... (i)
Now consider ∆ADC:
P and Q are the mid-points of CD and DA respectively.
By the Midpoint Theorem, PQ || AC and \( PQ = \frac{1}{2} AC \) ... (ii)
From (i) and (ii), we see that LM || PQ (since both are parallel to AC) and LM = PQ (since both are half of AC).
Since one pair of opposite sides of quadrilateral LMPQ is both parallel and equal, LMPQ is a parallelogram.
(iii) If it is also given that the diagonals AC and BD are equal, what further statement can be made about the parallelogram LMPQ?
If diagonals AC and BD are equal, then the quadrilateral ABCD can be a parallelogram or a rectangle.
From the Midpoint Theorem, we know that:
\( LM = \frac{1}{2} AC \)
\( MP = \frac{1}{2} BD \) (P and M are midpoints of CD and BC, so MP || BD and \( MP = \frac{1}{2} BD \))
\( PQ = \frac{1}{2} AC \)
\( QL = \frac{1}{2} BD \) (Q and L are midpoints of DA and AB, so QL || BD and \( QL = \frac{1}{2} BD \))
If AC = BD, then it follows that \( LM = MP = PQ = QL \).
Since LMPQ is a parallelogram with all its sides equal, LMPQ will be a rhombus.
In simple words: (i) The line LM connects the middle points of AB and BC. So, LM is parallel to AC and half as long as AC. (ii) We prove LMPQ is a parallelogram because LM and PQ are both parallel to AC and both half its length. This means LM is parallel and equal to PQ. Also, MP and QL are both parallel to BD and half its length, so they are parallel and equal too. (iii) If the diagonals AC and BD of the big quadrilateral are equal in length, then all sides of the inner parallelogram LMPQ will be equal. A parallelogram with equal sides is called a rhombus.

🎯 Exam Tip: This question tests the Midpoint Theorem and its converse. For part (iii), remember that equal diagonals in the original quadrilateral will make the midpoint quadrilateral have equal adjacent sides, turning it into a rhombus from a general parallelogram.

Question 14. ABCD is a parallelogram E is the mid-point of AB and F is the mid-point of CD. GH is any line that intersects AD, EF and BC in G, P and H respectively. Prove that GP = PH.
Answer:
Given: ABCD is a parallelogram. E is the mid-point of side AB, and F is the mid-point of side CD. GH is a line that cuts AD at G, EF at P, and BC at H.
To prove: GP = PH.
Proof:
Since E and F are the mid-points of AB and DC respectively, and ABCD is a parallelogram, it means that EF is parallel to AD and also parallel to BC. This forms parallel lines that are equally spaced. In such a setup, if a transversal line (like GH) cuts these parallel lines, it will divide them in equal segments if the parallel lines themselves are cut into equal parts by other transversals. The intercept theorem states that if three or more parallel lines make equal intercepts on one transversal, they will also make equal intercepts on any other transversal.
Here, because EF || AD || BC, and E and F are midpoints, the segment AD and BC are effectively "intercepted" by the line EF in a way that relates to the midpoints. The line GH is another transversal.
Given that EF || AD || BC, and E and F are mid-points, then the line EF is exactly midway between AD and BC. When the line GH crosses these three parallel lines (AD, EF, BC), it divides the segments proportionally. Since EF is exactly in the middle of AD and BC (due to E and F being mid-points of opposite sides of a parallelogram, making AEFD and EBCF parallelograms), then P, being the intersection on EF, must divide GH into equal parts, meaning GP = PH by the intercept theorem.
Therefore, GP = PH. Hence proved.
In simple words: Because E and F are in the middle of the parallelogram's sides, the line EF runs exactly in the middle of the parallelogram. If another line cuts through the top, middle, and bottom lines of the parallelogram, then the middle line will divide that cutting line into two equal pieces.

🎯 Exam Tip: When proving line segments are equal using parallel lines, always look for opportunities to apply the Mid-point Theorem or the Intercept Theorem. Clearly identify the parallel lines and the transversals.

Question 15. Prove that in a parallelogram, the lines joining a pair of opposite vertices to the mid-points of a pair of opposite sides trisect a diagonal.
Answer:
Given: In parallelogram ABCD, E is the mid-point of side AD, and F is the mid-point of side BC. The lines EB and DF are drawn, and they intersect the diagonal AC at points M and N respectively.
To prove: M and N trisect the diagonal AC, which means AM = MN = NC.
Proof:
Since ABCD is a parallelogram, its opposite sides are parallel and equal. So, AD || BC and AD = BC. Since E and F are the mid-points of AD and BC respectively, it means AE = ED and BF = FC. Also, AE || BF and AE = BF (because AD = BC). This makes the quadrilateral AEFB a parallelogram, which means that the line segment EB is parallel to AF. Similarly, since ED || FC and ED = FC, the quadrilateral ECDF is a parallelogram, which means the line segment DF is parallel to EC. Therefore, EB || DF.
Now, let's look at triangle ADN. E is the mid-point of side AD (given). The line segment EM is part of EB, and we know EB || DF. So, EM is parallel to DN. According to the converse of the Mid-point Theorem (also known as the intercept theorem extension), if a line passes through the mid-point of one side of a triangle and is parallel to another side, it bisects the third side. Therefore, M must be the mid-point of AN. This tells us that AM = MN (Equation i).
Next, consider triangle CBM. F is the mid-point of side BC (given). The line segment FN is part of DF, and we know DF || EB. So, FN is parallel to BM. Applying the converse of the Mid-point Theorem again, N must be the mid-point of MC. This means that CN = NM (Equation ii).
From both Equation (i) and Equation (ii), we can conclude that AM = MN = NC. This proves that the points M and N divide the diagonal AC into three equal parts, which means they trisect it. Hence proved.
In simple words: When you connect the corners of a parallelogram to the middle points of the opposite sides, these connecting lines cut the main diagonal into three equal parts.

🎯 Exam Tip: To prove trisection of a diagonal, break the problem into two smaller triangles. Apply the converse of the Mid-point Theorem twice, first in one triangle and then in the other, to show that the points divide the segments equally.

Question 16. In figure, E is the mid-point of side AD of a trapezium ABCD, with AB || DC. A line through E parallel to AB, meets BC in F. Show that F is the mid-point of BC.
Answer:
Given: ABCD is a trapezium, which means AB is parallel to DC. E is the mid-point of the side AD. A line is drawn through E that is parallel to AB, and this line meets the side BC at point F.
To prove: F is the mid-point of BC.
Proof:
First, let's draw a diagonal BD. Let this diagonal intersect the line EF at point O.
Now, consider triangle ADB. We are given that E is the mid-point of side AD. We also know that EF is parallel to AB (given). Since EO is a part of EF, it means EO is also parallel to AB.
According to the converse of the Mid-point Theorem, if a line passes through the mid-point of one side of a triangle and is parallel to another side, it must bisect the third side. Therefore, O must be the mid-point of the diagonal BD.
Next, consider triangle BCD. We have just proved that O is the mid-point of BD. We know that EF is parallel to AB (given), and AB is parallel to DC (given that ABCD is a trapezium). This means that EF is also parallel to DC. Since OF is a part of EF, it means OF is parallel to DC.
Again, applying the converse of the Mid-point Theorem in triangle BCD: a line (OF) passes through the mid-point (O) of one side (BD) and is parallel to another side (DC). Therefore, it must bisect the third side (BC). This means F is the mid-point of BC.
Thus, we have shown that F is the mid-point of BC. Hence proved.
In simple words: Imagine a shape with two parallel sides (a trapezium). If you start from the middle of one non-parallel side and draw a line parallel to the parallel sides, that line will automatically hit the exact middle of the other non-parallel side.

🎯 Exam Tip: When dealing with trapeziums and mid-points, drawing a diagonal is often the key. This helps you apply the Mid-point Theorem (or its converse) in two separate triangles to bridge across the trapezium.