OP Malhotra Class 9 Maths Solutions Chapter 8 Triangles Exercise 8 (A)

Question 1. In the figure, AB and CD bisect each other at K. Prove that AC = BD.
Answer:Given: Line segments AB and CD bisect each other at point K. This means K is the midpoint of both AB and CD. AC and BD are joined.
To prove: \( AC = BD \)
Proof: Consider the triangles \( \Delta AKC \) and \( \Delta BKD \).
We know that \( AK = BK \) (since K bisects AB, this is given).
Also, \( CK = DK \) (since K bisects CD, this is given).
The angles \( \angle AKC \) and \( \angle BKD \) are vertically opposite angles, so they are equal.
\( \implies \Delta AKC \cong \Delta BKD \) by the SAS (Side-Angle-Side) congruence rule.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies AC = BD \) (by c.p.c.t., which means corresponding parts of congruent triangles).
Therefore, it is proved that AC = BD.
In simple words: When two lines cut each other in half at one point, the opposite triangles formed are exactly the same size and shape. Because they are the same, their third sides must also be equal.

🎯 Exam Tip: When proving line segments equal, look for triangles that contain those segments and try to prove them congruent. Remember the conditions for SAS congruence: two sides and the included angle must be equal.

Question 2. In the figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove that DE || BC.
Answer:Given: In triangle ABC, sides BA and CA are extended. Point D is on the extension of BA such that \( BA = AD \). Point E is on the extension of CA such that \( CA = AE \). Line segment ED is drawn.
To prove: \( DE \parallel BC \)
Proof: Consider the triangles \( \Delta BAC \) and \( \Delta DAE \).
We are given that \( BA = AD \).
We are also given that \( CA = AE \).
The angles \( \angle BAC \) and \( \angle DAE \) are vertically opposite angles, so they are equal.
\( \implies \Delta BAC \cong \Delta DAE \) by the SAS (Side-Angle-Side) congruence rule.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies \angle ABC = \angle ADE \) (by c.p.c.t.).
These angles, \( \angle ABC \) and \( \angle ADE \), are alternate interior angles formed by the transversal BD intersecting lines DE and BC. When alternate interior angles are equal, the lines are parallel.
\( \implies DE \parallel BC \).
Hence, it is proved that DE is parallel to BC.
In simple words: We can show that the two triangles are the same. Because they are the same, certain angles must be equal. These equal angles are "alternate angles," which always means the lines are parallel.

🎯 Exam Tip: To prove lines parallel, look for pairs of equal alternate interior angles or corresponding angles. Proving triangles congruent is a common first step to establish these angle equalities.

Question 3. In the figure, ABCD is a rectangle; P is the mid-point of AB, Q and R are points in AD and BC respectively such that AQ = BR. Prove that PQ = PR.
Answer:Given: ABCD is a rectangle. P is the midpoint of side AB. Q is a point on AD and R is a point on BC such that \( AQ = BR \). Line segments PQ and PR are joined.
To prove: \( PQ = PR \)
Proof: Consider the triangles \( \Delta PAQ \) and \( \Delta PBR \).
We know that \( PA = PB \) because P is the midpoint of AB.
In a rectangle, all angles are 90 degrees. So, \( \angle A = 90^\circ \) and \( \angle B = 90^\circ \). Therefore, \( \angle A = \angle B \).
We are given that \( AQ = BR \).
\( \implies \Delta PAQ \cong \Delta PBR \) by the SAS (Side-Angle-Side) congruence rule.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies PQ = PR \) (by c.p.c.t.).
Hence, it is proved that PQ = PR.
In simple words: Since ABCD is a rectangle, angles A and B are 90 degrees. P is the middle of AB, so PA and PB are equal. With AQ equal to BR, the two small triangles at the top (PAQ and PBR) are exactly the same, which means PQ and PR are also equal.

🎯 Exam Tip: Remember that rectangles have 90-degree angles at each corner and opposite sides are equal. The midpoint divides a side into two equal halves. These properties are crucial for congruence proofs.

Question 4. In the figure, OA = OB, OC = OD, ∠AOB = ∠COD, prove that AC = BD.
Answer:Given: In the figure, \( OA = OB \), \( OC = OD \), and \( \angle AOB = \angle COD \).
To prove: \( AC = BD \)
Proof: We are given that \( \angle AOB = \angle COD \).
Let's add \( \angle COB \) to both sides of this equality.
\( \implies \angle AOB + \angle COB = \angle COD + \angle COB \)
From the figure, \( \angle AOB + \angle COB \) makes the angle \( \angle AOC \).
And \( \angle COD + \angle COB \) makes the angle \( \angle BOD \).
\( \implies \angle AOC = \angle BOD \).
Now consider the triangles \( \Delta OAC \) and \( \Delta ODB \).
We are given \( OA = OB \).
We are given \( OC = OD \).
We have just proved that \( \angle AOC = \angle BOD \).
\( \implies \Delta OAC \cong \Delta ODB \) by the SAS (Side-Angle-Side) congruence rule. This is because we have two sides and the included angle equal.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies AC = BD \) (by c.p.c.t.).
Hence, it is proved that AC = BD.
In simple words: First, we show that if two small angles are equal, then adding the same middle angle to both will make two larger angles equal. Then, using the given equal sides and these newly equal angles, we prove that the two main triangles are the same, so their third sides must also be equal.

🎯 Exam Tip: When given an equality between small angles that share a common region, try adding the common angle to both sides to form larger angles. This technique is often useful for SAS congruence proofs.

Question 5. In figure. prove that As OAM and OBN are congruent and hence prove that AM = BN.
Answer:Given: In the figure, \( OA = OB \). AM is perpendicular to XY, and BN is perpendicular to XY.
To prove: (i) \( \Delta OAM \cong \Delta OBN \) (ii) \( AM = BN \)
Proof:
(i) Consider the triangles \( \Delta OAM \) and \( \Delta OBN \).
We are given that \( OA = OB \).
Since AM is perpendicular to XY, \( \angle OMA = 90^\circ \).
Since BN is perpendicular to XY, \( \angle ONB = 90^\circ \).
So, \( \angle OMA = \angle ONB = 90^\circ \).
The angles \( \angle AOM \) and \( \angle BON \) are vertically opposite angles, so they are equal.
\( \implies \Delta OAM \cong \Delta OBN \) by the AAS (Angle-Angle-Side) congruence rule. This rule applies because we have two pairs of equal angles and one pair of equal non-included sides.
(ii) Since the triangles \( \Delta OAM \) and \( \Delta OBN \) are congruent, their corresponding parts are equal.
\( \implies AM = BN \) (by c.p.c.t.).
Hence, both parts are proved.
In simple words: We know that two sides from the center (O) to A and B are equal. We also have two right angles and two vertically opposite angles. This is enough to show that the two triangles are identical, which means their heights (AM and BN) must also be equal.

🎯 Exam Tip: When dealing with perpendiculars, remember that they form right angles. Also, vertically opposite angles are always equal. These two facts, combined with a given equal side, often point towards AAS congruence.

Question 6. In Figure, ∠XYZ is bisected by YP. L is any point in YP and MLN is perpendicular to YP. Prove that LM = LN.
Answer:Given: In the figure, the angle \( \angle XYZ \) is bisected by the line YP. L is any point on YP. MLN is a line segment perpendicular to YP.
To prove: \( LM = LN \)
Proof: Consider the triangles \( \Delta YML \) and \( \Delta YNL \).
The line segment YL is common to both triangles. So, \( YL = YL \).
Since MLN is perpendicular to YP, the angles \( \angle YLM \) and \( \angle YLN \) are both 90 degrees.
So, \( \angle YLM = \angle YLN = 90^\circ \).
We are given that YP bisects \( \angle XYZ \). This means that YP divides \( \angle XYZ \) into two equal angles.
Therefore, \( \angle MYL = \angle NYL \).
\( \implies \Delta YML \cong \Delta YNL \) by the ASA (Angle-Side-Angle) congruence rule. We have two pairs of equal angles and the included side YL equal.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies LM = LN \) (by c.p.c.t.).
Hence, it is proved that LM = LN.
In simple words: When a line cuts an angle exactly in half, and you draw a straight line from a point on the bisector that hits the original lines at a right angle, then the two pieces of that straight line will be equal in length. This is because the two triangles formed are exactly the same.

🎯 Exam Tip: An angle bisector divides an angle into two equal parts. A perpendicular line forms a 90-degree angle. Look for these properties to establish angle and side equalities for ASA congruence.

Question 7. In figure, X is any point within a square ABCD. On AX a square AXYZ is described. Prove that BX = DZ.
Answer:Given: ABCD is a square. X is any point inside the square ABCD. On AX, another square AXYZ is drawn. BX and DZ are joined.
To prove: \( BX = DZ \)
Proof: Consider the triangles \( \Delta ABX \) and \( \Delta ADZ \).
Since ABCD is a square, all its sides are equal. So, \( AB = AD \).
Since AXYZ is a square, all its sides are equal. So, \( AX = AZ \).
We know that \( \angle DAB = 90^\circ \) (angle of square ABCD).
We also know that \( \angle XAZ = 90^\circ \) (angle of square AXYZ).
We can write \( \angle BAX \) as \( \angle DAB - \angle DAX \).
We can write \( \angle DAZ \) as \( \angle XAZ - \angle DAX \).
Since \( \angle DAB = 90^\circ \) and \( \angle XAZ = 90^\circ \), and \( \angle DAX \) is common to both subtractions:
\( \angle BAX = 90^\circ - \angle DAX \)
\( \angle DAZ = 90^\circ - \angle DAX \)
\( \implies \angle BAX = \angle DAZ \).
Now we have:
\( AB = AD \) (sides of square ABCD)
\( AX = AZ \) (sides of square AXYZ)
\( \angle BAX = \angle DAZ \) (proved above)
\( \implies \Delta ABX \cong \Delta ADZ \) by the SAS (Side-Angle-Side) congruence rule.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies BX = DZ \) (by c.p.c.t.).
Hence, it is proved that BX = DZ.
In simple words: Both ABCD and AXYZ are squares, so they have equal sides and 90-degree angles. By looking at the angles around point A, we can see that the angle in triangle ABX is the same as the angle in triangle ADZ. Because of this, and the equal sides, the two triangles are identical, which means BX must be equal to DZ.

🎯 Exam Tip: When working with squares, remember that all sides are equal and all angles are 90 degrees. If an angle is shared or can be expressed as a common angle subtracted from 90 degrees, it can help establish angle equality for congruence.

Question 8. In figure, prove that AP bisects ∠BAC.
Answer:Given: In the figure, P is a point inside \( \angle BAC \). PM is perpendicular to AB, and PN is perpendicular to AC. We are given that \( PM = PN \).
To prove: AP bisects \( \angle BAC \).
Proof: Consider the two right-angled triangles, \( \Delta PAM \) and \( \Delta PAN \).
We are given that \( PM = PN \).
The side AP is common to both triangles. So, \( AP = AP \). (This is the hypotenuse for both right-angled triangles).
Since PM is perpendicular to AB, \( \angle PMA = 90^\circ \).
Since PN is perpendicular to AC, \( \angle PNA = 90^\circ \).
\( \implies \Delta PAM \cong \Delta PAN \) by the RHS (Right angle-Hypotenuse-Side) congruence rule.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies \angle PAM = \angle PAN \) (by c.p.c.t.).
Since AP divides \( \angle BAC \) into two equal angles, AP bisects \( \angle BAC \).
Hence, it is proved.
In simple words: If a point inside an angle is the same distance from both arms of the angle, and we connect that point to the corner of the angle, then that connecting line will split the angle exactly in half. This is because the two right-angled triangles formed are identical.

🎯 Exam Tip: The RHS congruence rule is specifically for right-angled triangles and requires the hypotenuse and one side to be equal. When dealing with angle bisectors, often the goal is to show that angles formed are equal.

Question 9. In figure, if PR = PS, PT = PQ, ∠PTS = ∠PQR = 90°, prove that RQ = ST and RT = SQ.
Answer:Given: In the figure, \( PR = PS \), \( PT = PQ \). Also, \( \angle PTS = 90^\circ \) and \( \angle PQR = 90^\circ \).
To prove: (i) \( RQ = ST \) (ii) \( RT = SQ \)
Proof:
(i) Consider the two right-angled triangles \( \Delta PQR \) and \( \Delta PTS \).
We are given that \( PQ = PT \). (This is one side).
We are given that \( PR = PS \). (This is the hypotenuse, opposite the 90-degree angle).
We know that \( \angle PQR = 90^\circ \) and \( \angle PTS = 90^\circ \).
\( \implies \Delta PQR \cong \Delta PTS \) by the RHS (Right angle-Hypotenuse-Side) congruence rule.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies RQ = ST \) (by c.p.c.t.).
(ii) We are given \( PR = PS \) and \( PT = PQ \).
We can subtract the smaller segment from the larger segment.
Subtracting \( PT \) from \( PR \), we get \( PR - PT = RT \).
Subtracting \( PQ \) from \( PS \), we get \( PS - PQ = SQ \).
Since \( PR = PS \) and \( PT = PQ \), subtracting equal quantities from equal quantities means the results are also equal.
\( \implies PR - PT = PS - PQ \)
\( \implies RT = SQ \).
Hence, both parts are proved.
In simple words: First, by looking at the two right-angled triangles, we can see they are identical because they have matching hypotenuses and one other side. This makes the remaining sides, RQ and ST, equal. Second, if we start with two equal lengths and remove two other equal lengths from them, the leftover parts (RT and SQ) must also be equal.

🎯 Exam Tip: For problems with multiple parts, the result of the first part (like triangle congruence) often helps solve the second part. Also, remember that subtracting equal quantities from equal quantities yields equal results.

Question 10. In figure, PQRS is a square. Arc AB is drawn with centre P and any radius less than PR, cutting SR at A and RQ at B. Prove that AS = BQ.
Answer:Given: PQRS is a square. An arc AB is drawn with center P and a radius smaller than PR. This arc cuts SR at A and RQ at B. So, PA and PB are radii of the same arc.
To prove: \( AS = BQ \)
Proof: Consider the two right-angled triangles \( \Delta PSA \) and \( \Delta PQB \).
Since PQRS is a square, all its sides are equal. So, \( PS = PQ \).
PA and PB are radii of the same arc with center P. Therefore, \( PA = PB \). (This is the hypotenuse for both right-angled triangles because angles S and Q are 90 degrees).
Since PQRS is a square, \( \angle S = 90^\circ \) and \( \angle Q = 90^\circ \).
\( \implies \Delta PSA \cong \Delta PQB \) by the RHS (Right angle-Hypotenuse-Side) congruence rule.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies AS = BQ \) (by c.p.c.t.).
Hence, it is proved that AS = BQ.
In simple words: Since PQRS is a square, its corners P, Q, R, S are 90 degrees, and sides PS and PQ are equal. Because PA and PB are radii of the same arc from point P, they are also equal. This makes the two right-angled triangles, PSA and PQB, identical. Therefore, their third sides, AS and BQ, must be equal.

🎯 Exam Tip: In squares, all angles are 90 degrees, and adjacent sides are equal. Radii of the same arc are always equal. These are essential facts to use for RHS congruence proofs.

Question 11. In figure, if LM = MN, QM = MR and ∠MLQ =
Answer:Given: In the figure, \( LM = MN \), \( QM = MR \). Also, \( \angle MLQ = 90^\circ \) and \( \angle MNR = 90^\circ \).
To prove: \( PQ = PR \)
Proof: First, consider the two right-angled triangles \( \Delta QLM \) and \( \Delta RNM \).
We are given that \( LM = MN \).
We are given that \( QM = MR \). (This is the hypotenuse for both right-angled triangles).
We are given that \( \angle MLQ = 90^\circ \) and \( \angle MNR = 90^\circ \).
\( \implies \Delta QLM \cong \Delta RNM \) by the RHS (Right angle-Hypotenuse-Side) congruence rule.
Since these two triangles are congruent, their corresponding parts are equal.
\( \implies \angle LQM = \angle NRM \) (by c.p.c.t.). This means \( \angle Q = \angle R \).
Now, consider the larger triangle \( \Delta PQR \).
We have just proved that \( \angle Q = \angle R \).
In any triangle, the sides opposite to equal angles are equal.
The side opposite to \( \angle Q \) is PR.
The side opposite to \( \angle R \) is PQ.
\( \implies PR = PQ \) (sides opposite equal angles).
Therefore, \( PQ = PR \).
Hence, it is proved.
In simple words: We first show that the two small right-angled triangles at the bottom are identical because they have matching hypotenuses and one side. This makes the angles at Q and R equal. Since angles Q and R in the large triangle PQR are equal, the sides opposite them (PR and PQ) must also be equal.

🎯 Exam Tip: This problem demonstrates a common two-step proof strategy: first prove smaller triangles congruent to establish equality of angles or sides, then use those equalities in a larger figure to prove the final statement.

Question 12. In figure, ∆ABC is right-angled at B. ACDE and BCGF are squares. Prove that (i) ΔBCD ≅ ΔACG (ii) AG = BD
Answer:Given: \( \Delta ABC \) is a right-angled triangle with the right angle at B. ACDE is a square drawn on side AC, and BCGF is a square drawn on side BC. AG and BD are joined.
To prove: (i) \( \Delta BCD \cong \Delta ACG \) (ii) \( AG = BD \)
Proof:
(i) Consider the squares ACDE and BCGF.
Since ACDE is a square, \( AC = CD \) and \( \angle ACD = 90^\circ \).
Since BCGF is a square, \( BC = CG \) and \( \angle BCG = 90^\circ \).
Now, let's consider the angles \( \angle BCD \) and \( \angle ACG \).
We have \( \angle ACD = 90^\circ \) and \( \angle BCG = 90^\circ \).
We can add \( \angle ACB \) to both \( \angle ACD \) and \( \angle BCG \).
\( \angle ACD + \angle ACB = 90^\circ + \angle ACB \)
\( \angle BCG + \angle ACB = 90^\circ + \angle ACB \)
\( \implies \angle BCD = \angle ACG \).
Now, consider the triangles \( \Delta BCD \) and \( \Delta ACG \).
We have \( BC = CG \) (sides of square BCGF).
We have \( CD = AC \) (sides of square ACDE).
We have \( \angle BCD = \angle ACG \) (proved above).
\( \implies \Delta BCD \cong \Delta ACG \) by the SAS (Side-Angle-Side) congruence rule.
(ii) Since \( \Delta BCD \cong \Delta ACG \) (proved in part i), their corresponding parts are equal.
\( \implies BD = AG \) (by c.p.c.t.).
Hence, both parts are proved.
In simple words: First, because squares have 90-degree corners, and we add the common angle ACB to two 90-degree angles, the larger angles BCD and ACG become equal. Then, using the equal sides of the squares and these equal angles, we show that triangle BCD is exactly the same as triangle ACG. Since these triangles are identical, their third sides, BD and AG, must also be equal.

🎯 Exam Tip: When squares are constructed on the sides of a triangle, look for combinations of angles and sides that allow you to prove congruence. Adding a common angle to two equal angles (like 90 degrees) is a frequent strategy to show that new, larger angles are equal.

Question 13. Prove that the medians of an equilateral triangle are equal.
Answer:Given: \( \Delta ABC \) is an equilateral triangle. AD, BE, and CF are its medians.
To prove: \( AD = BE = CF \)
Proof: In an equilateral triangle, all sides are equal and all angles are 60 degrees. So, \( AB = BC = CA \) and \( \angle A = \angle B = \angle C = 60^\circ \).
Since D, E, F are midpoints, we have:
\( CD = DB = \frac{1}{2} BC \)
\( AE = EC = \frac{1}{2} AC \)
\( AF = FB = \frac{1}{2} AB \)
Since \( AB = BC = CA \), it follows that \( AE = EC = CD = DB = AF = FB \).
Let's consider the triangles \( \Delta EBC \) and \( \Delta FCB \).
\( BC = CB \) (common side).
\( EC = FB \) (since they are halves of equal sides AC and AB respectively).
\( \angle ECB = \angle FBC = 60^\circ \) (angles of an equilateral triangle, i.e., \( \angle C = \angle B \)).
\( \implies \Delta EBC \cong \Delta FCB \) by the SAS (Side-Angle-Side) congruence rule.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies BE = CF \) (by c.p.c.t.). This is equation (i).
Similarly, we can prove that \( \Delta FCB \cong \Delta DAC \).
\( BC = AC \) (sides of equilateral triangle)
\( FB = DC \) (halves of equal sides AB and BC)
\( \angle B = \angle C \) (angles of equilateral triangle)
\( \implies \Delta FCB \cong \Delta DAC \) by the SAS congruence rule.
\( \implies CF = AD \) (by c.p.c.t.). This is equation (ii).
From (i) and (ii), we can conclude that \( AD = BE = CF \).
Hence, the medians of an equilateral triangle are equal.
In simple words: An equilateral triangle has all sides and all angles equal. When you draw lines from each corner to the middle of the opposite side (these are called medians), they cut the triangle into smaller, identical parts. By proving these smaller triangles are the same, we can show that all three median lines must be equal in length.

🎯 Exam Tip: For proofs involving equilateral triangles, always start by stating the properties: all sides are equal, all angles are 60 degrees. This provides the necessary equalities for side and angle comparisons in congruence proofs.

Question 14. In figure, AC = DE, ∠ACB =∠EDF and BD = FC. Prove that AB = EF.
Answer:Given: In the figure, \( AC = DE \), \( \angle ACB = \angle EDF \), and \( BD = FC \).
To prove: \( AB = EF \)
Proof: We are given that \( BD = FC \).
Let's add the common segment DC to both sides of this equation.
\( BD + DC = FC + DC \)
From the figure, \( BD + DC \) forms the side BC.
And \( FC + DC \) forms the side DF.
\( \implies BC = DF \).
Now, consider the triangles \( \Delta ABC \) and \( \Delta EDF \).
We are given that \( AC = DE \).
We have just proved that \( BC = DF \).
We are given that \( \angle ACB = \angle EDF \).
\( \implies \Delta ABC \cong \Delta EDF \) by the SAS (Side-Angle-Side) congruence rule. This is because we have two pairs of equal sides and the included angle between them is also equal.
Since the triangles are congruent, their corresponding parts are equal.
\( \implies AB = EF \) (by c.p.c.t.).
Hence, it is proved that AB = EF.
In simple words: First, we use the given information to show that the entire base BC is equal to DF by adding the common segment DC. Then, since we know two sides and the angle between them are equal in both triangles, we can prove that triangle ABC and triangle EDF are identical. This means their remaining sides, AB and EF, must also be equal.

🎯 Exam Tip: When parts of lines are given as equal, check if adding a common segment can make larger segments equal. This is a crucial step in many congruence proofs that involve line segments with overlapping parts.

Question 15. In figure, AC = AE, AB = AD and ∠BAD =∠EAC. Prove that BC = DE.
Answer: