RBSE Solutions Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Exercise 7.3

Question 1. ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
(i) ∆ABD = ∆ACD
(ii) ∆ABP = ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.

Answer:
(i) Consider triangles ∆ABD and ∆ACD:
AB = AC (Given, as ∆ABC is isosceles)
BD = DC (Given, as ∆DBC is isosceles)
AD = AD (Common side)
Thus, ∆ABD \( \cong \) ∆ACD (by SSS congruency rule).

(ii) Consider triangles ∆ABP and ∆ACP:
AB = AC (Given)
From (i), ∆ABD \( \cong \) ∆ACD
\( \implies \) ∠BAD = ∠CAD (by c.p.c.t.)
\( \implies \) ∠BAP = ∠CAP
AP = AP (Common side)
Thus, ∆ABP \( \cong \) ∆ACP (by SAS congruency rule). This means that if two sides and the included angle of one triangle are equal to the corresponding parts of another triangle, they are congruent.

(iii) AP bisects ∠A as well as ∠D:
Since ∆ABP \( \cong \) ∆ACP (from part ii), we know ∠BAP = ∠CAP (by c.p.c.t). This proves AP bisects ∠A.
Now, consider triangles ∆BDP and ∆CDP:
BD = DC (Given)
DP = DP (Common side)
BP = CP (Since ∆ABP \( \cong \) ∆ACP from part (ii), by c.p.c.t.)
Thus, ∆BDP \( \cong \) ∆CDP (by SSS congruency rule).
Since ∆BDP \( \cong \) ∆CDP, we have ∠BDP = ∠CDP (by c.p.c.t). This proves AP (or DP) bisects ∠D.

(iv) AP is the perpendicular bisector of BC:
From part (iii), since ∆BDP \( \cong \) ∆CDP, we have BP = CP (by c.p.c.t). This means P is the midpoint of BC, so AP bisects BC.
Also from ∆BDP \( \cong \) ∆CDP, we have ∠BPD = ∠CPD (by c.p.c.t).
Since ∠BPD and ∠CPD form a linear pair, their sum is 180°.
∠BPD + ∠CPD = 180°
\( \implies \) ∠BPD + ∠BPD = 180°
\( \implies \) 2∠BPD = 180°
\( \implies \) ∠BPD = 90°
Since AP is perpendicular to BC (∠BPD = 90°) and bisects BC (BP = CP), AP is the perpendicular bisector of BC.
In simple words: We first show that the two smaller triangles formed by AD on the base BC are identical using SSS rule. Then, using this, we prove two other triangles (ABP and ACP) are identical by SAS rule, which helps confirm AP divides angle A equally. Next, we show that AP also divides angle D equally and cuts the base BC exactly in half at a 90-degree angle, meaning it is a perpendicular bisector.

🎯 Exam Tip: Always break down complex congruency proofs into smaller, manageable steps. Clearly state the congruency rule (SSS, SAS, RHS, ASA, AAS) used at each step and the corresponding parts (c.p.c.t).

Question 2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.

Answer:
Given: ∆ABC is an isosceles triangle with AB = AC. AD is an altitude to BC, which means AD \( \perp \) BC. Therefore, ∠ADB = ∠ADC = 90°.

(i) To show AD bisects BC (i.e., BD = DC):
Consider the right-angled triangles ∆ABD and ∆ACD:
∠ADB = ∠ADC = 90° (Given, as AD is an altitude)
Hypotenuse AB = Hypotenuse AC (Given)
AD = AD (Common side)
Thus, ∆ABD \( \cong \) ∆ACD (by RHS congruency rule). This rule states that if the hypotenuse and one side of a right-angled triangle are equal to the corresponding parts of another, the triangles are congruent.
Since the triangles are congruent, their corresponding parts are equal:
BD = DC (by c.p.c.t.)
Hence, AD bisects BC.

(ii) To show AD bisects ∠A (i.e., ∠BAD = ∠CAD):
Since ∆ABD \( \cong \) ∆ACD (from part i), their corresponding parts are equal:
∠BAD = ∠CAD (by c.p.c.t.)
Hence, AD bisects ∠A.
In simple words: When you draw a line from the top corner of an isosceles triangle straight down to the middle of the opposite side (making a 90-degree angle), that line not only splits the bottom side into two equal parts but also divides the top angle into two equal angles.

🎯 Exam Tip: For problems involving right-angled triangles and equal hypotenuses, the RHS (Right-angle, Hypotenuse, Side) congruence rule is often the most direct method. Remember to clearly identify the right angle, hypotenuse, and one side.

Question 3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to side PQ and QR and median PN of ∆PQR (see figure). Show that
(i) ∆ABM = ∆PQN
(ii) ∆ABC = ∆PQR

Answer:
Given: In ∆ABC and ∆PQR, we have:
AB = PQ
BC = QR
Median AM = Median PN

(i) To show ∆ABM \( \cong \) ∆PQN:
Since AM is the median of ∆ABC, M is the midpoint of BC. So, BM = \( \frac{1}{2} \)BC.
Since PN is the median of ∆PQR, N is the midpoint of QR. So, QN = \( \frac{1}{2} \)QR.
We are given BC = QR.
Multiplying both sides by \( \frac{1}{2} \), we get \( \frac{1}{2} \)BC = \( \frac{1}{2} \)QR.
\( \implies \) BM = QN.
Now, consider triangles ∆ABM and ∆PQN:
AB = PQ (Given)
AM = PN (Given)
BM = QN (Proved above)
Thus, ∆ABM \( \cong \) ∆PQN (by SSS congruency rule). This means that if all three sides of one triangle are equal to the corresponding three sides of another, the triangles are congruent.

(ii) To show ∆ABC \( \cong \) ∆PQR:
Since ∆ABM \( \cong \) ∆PQN (from part i), their corresponding parts are equal:
∠B = ∠Q (by c.p.c.t.)
Now, consider triangles ∆ABC and ∆PQR:
AB = PQ (Given)
∠B = ∠Q (Proved above)
BC = QR (Given)
Thus, ∆ABC \( \cong \) ∆PQR (by SAS congruency rule).
In simple words: First, we prove that the smaller triangles formed by the medians are identical by comparing their three sides. Because these smaller triangles are identical, we know that one of their angles (angle B and angle Q) must also be the same. Then, using this equal angle along with the two given equal sides, we prove that the two main, larger triangles (ABC and PQR) are also identical.

🎯 Exam Tip: When medians are involved, remember that they bisect the opposite side. This fact helps establish side equality for SSS or SAS congruence. Always use the results of previous congruencies (like c.p.c.t.) to prove subsequent parts.

Question 4. BE and CF are two equal altitudes of ∆ABC. By using RHS congruency rule, prove that ∆ABC is an isosceles triangle.

Answer:
Given: In ∆ABC, BE and CF are altitudes, so BE \( \perp \) AC and CF \( \perp \) AB. We are also given that the altitudes are equal, so BE = CF.

To prove: ∆ABC is an isosceles triangle, meaning AB = AC or ∠B = ∠C.
Consider the right-angled triangles ∆BFC and ∆CEB:
∠BFC = 90° (Since CF is an altitude)
∠CEB = 90° (Since BE is an altitude)
Hypotenuse BC = Hypotenuse BC (Common side)
CF = BE (Given that altitudes are equal)
Thus, ∆BFC \( \cong \) ∆CEB (by RHS congruency rule). The RHS rule is very useful for proving congruence in right-angled triangles.
Since the triangles are congruent, their corresponding parts are equal:
∠FBC = ∠ECB (by c.p.c.t.)
\( \implies \) ∠B = ∠C
In ∆ABC, since ∠B = ∠C, the sides opposite these equal angles must also be equal.
Therefore, AC = AB.
Hence, ∆ABC is an isosceles triangle.
In simple words: If a triangle has two equal height lines (altitudes), we can use this to show that two smaller right-angled triangles within it are identical. Because these small triangles are identical, it means the main triangle has two equal angles at its base, which then proves that the triangle itself must have two equal sides, making it an isosceles triangle.

🎯 Exam Tip: To prove a triangle is isosceles, a common strategy is to prove that two of its angles or two of its sides are equal. Using RHS congruence with altitudes is a powerful method for this, as it directly involves right angles and often a common hypotenuse.

Question 5. ∆ABC is an isosceles triangle with AB = AC. Draw AP \( \perp \) BC to show that ∠B = ∠C.

Answer:
Given: ∆ABC is an isosceles triangle with AB = AC.

Construction: Draw a line segment AP from vertex A perpendicular to side BC, such that P lies on BC.
So, AP \( \perp \) BC, which means ∠APB = ∠APC = 90°.

To show: ∠B = ∠C.
Consider the right-angled triangles ∆ABP and ∆ACP:
∠APB = ∠APC = 90° (By construction)
Hypotenuse AB = Hypotenuse AC (Given)
AP = AP (Common side)
Thus, ∆ABP \( \cong \) ∆ACP (by RHS congruency rule). This rule helps establish congruence when a right angle, hypotenuse, and one side are equal.
Since the triangles are congruent, their corresponding parts are equal:
∠B = ∠C (by c.p.c.t.)
Hence, we have shown that in an isosceles triangle, the angles opposite to the equal sides are equal.
In simple words: When you have a triangle with two equal sides (isosceles), if you draw a line from the top corner straight down to the base at a right angle, you create two identical right-angled triangles. Because these smaller triangles are exactly the same, their matching angles at the base (angle B and angle C) must also be equal.

🎯 Exam Tip: When asked to prove angle or side equality in an isosceles triangle, drawing an altitude or a median can create congruent triangles that make the proof straightforward using rules like RHS or SSS.