GSEB Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.3

Question 1. ∆ABC and ADBC 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 1. ΔABD = ΔACD 2. ΔABP = ΔACP 3. AP bisects ∠A as well as ∠D 4. AP is the perpendicular bisector of BC.

Answer:
1. In triangle ABD and triangle ACD, we see AB equals AC because it's given that triangle ABC is an isosceles triangle. Also, BD equals CD since triangle DBC is an isosceles triangle. The side AD is common to both triangles. Therefore, triangle ABD is congruent to triangle ACD by the SSS (Side-Side-Side) congruency rule.
In simple words: We compare two triangles, ABD and ACD. Their sides are equal (AB=AC, BD=CD, AD=AD). So, the triangles are exactly the same shape and size.
2. In triangles ABP and ACP, we know AB equals AC because it's given. The angle BAP equals angle CAP, which we learned from CPCT (Corresponding Parts of Congruent Triangles) because we already proved triangle ABD is congruent to triangle ACD. The side AP is common to both triangles. Thus, triangle ABP is congruent to triangle ACP by the SAS (Side-Angle-Side) congruency rule.
In simple words: We look at triangles ABP and ACP. They have equal sides (AB=AC, AP=AP) and the angle between those sides is also equal. This means the triangles are congruent.
3. Since triangle ABP is congruent to triangle ACP (as proved in part (ii)), we can say that BP equals CP (from CPCT) and angle BAP equals angle CAP (from CPCT). This demonstrates that AP divides angle A into two equal halves. Next, consider triangles BDP and CDP. We are given that BD equals CD because triangle BDC is an isosceles triangle. We also know BP equals CP, as proved earlier in part (ii). The side DP is common to both triangles. Therefore, triangles BDP and CDP are congruent by the SSS (Side-Side-Side) congruency rule. From this, angle BDP equals angle CDP (by CPCT), which means AP also divides angle D into two equal halves.
In simple words: Because triangles ABP and ACP are the same, AP splits angle A into two equal parts. Similarly, because triangles BDP and CDP are the same, AP also splits angle D into two equal parts.
4. Since triangle BDP is congruent to triangle CDP (as proved above), we know that BP equals CP (by CPCT) and angle BPD equals angle CPD (by CPCT). We also know that angle BPD and angle CPD form a linear pair, so their sum is 180 degrees. Since both angles are equal, we can write 2 times angle BPD equals 180 degrees, which simplifies to angle BPD equals 90 degrees. This result shows that AP is the perpendicular bisector of BC, meaning it cuts BC exactly in half and at a right angle.
In simple words: We know the triangles BDP and CDP are identical. This makes BP equal to CP and angles BPD and CPD equal. Since these two angles add up to 180 degrees and are equal, each must be 90 degrees. So, AP cuts BC in half and at a right angle.

Exam Tip: For problems involving congruency, clearly state the congruency rule (SSS, SAS, ASA, RHS) and CPCT for corresponding parts.

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:
(i) Consider triangles ADB and ADC. We know that angle ADB equals angle ADC, both being 90 degrees, since AD is an altitude. Also, side AB equals side AC because triangle ABC is an isosceles triangle. The side AD is common to both triangles. Therefore, triangle ABD is congruent to triangle ACD by the RHS (Right angle-Hypotenuse-Side) congruency rule. This means that BD equals CD (by CPCT), which proves that AD divides BC into two equal segments.
In simple words: In the triangles ADB and ADC, both have a right angle. The longest side (hypotenuse) AB is equal to AC, and side AD is common. So, the triangles are congruent. This means AD cuts BC into two equal parts.
(ii) Since triangle ADB is congruent to triangle ADC (as proved in part (i)), it follows that angle BAD equals angle CAD (by CPCT). This clearly shows that AD divides angle A into two equal halves.
In simple words: Since triangles ADB and ADC are congruent, angle BAD is equal to angle CAD. This shows that AD cuts angle A into two equal angles.

Exam Tip: When dealing with isosceles triangles and altitudes, remember that the altitude to the base also acts as a median and an angle bisector from the vertex.

Question 3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see figure). Show that
(i) ΔABM = ΔPQN
(ii) ΔABC = ΔPQR.
Answer:

Answer:
Given two triangles, ABC and PQR, with sides AB equal to PQ, BC equal to QR, and median AM equal to median PN.
(i) For part (i), let's consider triangles ABM and PQN. We know AB equals PQ (given). We also know BC equals QR (given), so half of BC will equal half of QR. Since AM and PN are medians, they bisect the opposite sides, meaning BM equals QN. Also, AM equals PN (given). Therefore, triangle ABM is congruent to triangle PQN by the SSS (Side-Side-Side) congruency rule. As a result, angle ABM equals angle PQN (by CPCT),
\( \implies \) which implies that angle ABC equals angle PQR.
In simple words: We compare triangles ABM and PQN. All their sides are equal (AB=PQ, BM=QN, AM=PN). So, these two triangles are exactly the same. Because they're the same, angle ABM is equal to angle PQN, which means angle ABC is equal to angle PQR.
(ii) For part (ii), let's examine triangles ABC and PQR. We are given that AB equals PQ. We also know that angle ABC equals angle PQR, as we proved this in part (i). Furthermore, BC equals QR (given). Therefore, triangle ABC is congruent to triangle PQR by the SAS (Side-Angle-Side) congruency rule.
In simple words: Now we compare the bigger triangles, ABC and PQR. We know side AB is equal to PQ, side BC is equal to QR, and the angle between them (angle ABC and angle PQR) is also equal. This makes the two large triangles congruent.

Exam Tip: Always clearly state the given information, what needs to be proved, and the construction steps if any, before beginning the proof. This makes your argument clear and logical.

Question 4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Answer:

Answer:
Given that BE and CF are two equal altitudes of triangle ABC. We need to prove that triangle ABC is an isosceles triangle. Let's look at triangles BEC and CFB. We know that angle BEC equals angle CFB, both being 90 degrees, since BE and CF are altitudes. The side BC is common to both triangles. We are also given that BE equals CF. Therefore, triangle BEC is congruent to triangle CFB by the RHS (Right angle-Hypotenuse-Side) congruency rule. From this, angle BCF equals angle CBE (by CPCT). Since these angles are equal, the sides opposite to them, AB and AC, must also be equal. Thus, triangle ABC is an isosceles triangle.
In simple words: We compare triangles BEC and CFB. Both have a right angle, share side BC, and their altitudes BE and CF are equal. This makes them congruent. Because they are congruent, the angles opposite the equal altitudes are equal, which means the sides AB and AC are also equal, making triangle ABC isosceles.

Exam Tip: When using the RHS congruence rule, ensure you correctly identify the right angle, hypotenuse, and one corresponding side in both triangles.

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

Answer:
Given that ABC is an isosceles triangle where AB equals AC. We need to show that angle B equals angle C. First, we construct a line AP such that AP is perpendicular to BC. Now, consider the right-angled triangles APB and APC. Angle APB equals angle APC, both being 90 degrees. We are given that AB equals AC. The side AP is common to both triangles. Therefore, triangle APB is congruent to triangle APC by the RHS (Right angle-Hypotenuse-Side) congruency rule. Consequently, angle ABP equals angle ACP (by CPCT),
\( \implies \) which means angle B equals angle C.
In simple words: We have an isosceles triangle ABC with AB=AC. Draw a line AP that meets BC at a right angle. Now compare triangles APB and APC. They both have a right angle, share side AP, and their hypotenuses AB and AC are equal. So, the triangles are congruent. This means angle B and angle C are equal.

Exam Tip: This proof is fundamental for understanding properties of isosceles triangles. Remember that the altitude from the vertex angle of an isosceles triangle bisects the base and the vertex angle.