knowledge-base/class_9/ch7/0142_geometry.md

---
source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch7
exercise: ex7.3
difficulty: intermediate
---

Class 9 | Ch7: Triangles | Exercise 7.3

Q1. △ABC and △DBC are two isosceles triangles on same base BC and vertices A and D on same side of BC. AD is extended to intersect BC at P. Show:
(i) △ABD≅△ACD (ii) △ABP≅△ACP (iii) AP bisects ∠A and ∠D (iv) AP is perpendicular bisector of BC.
Answer:
(i) AB=AC, DB=DC, AD=AD → △ABD≅△ACD by SSS. ✅
(ii) AB=AC, BP=CP (from i, CPCT), AP=AP → △ABP≅△ACP by SSS. ✅
(iii) ∠BAP=∠CAP (CPCT from ii) → AP bisects ∠A. ∠BDP=∠CDP (CPCT from i) → AP bisects ∠D. ✅
(iv) BP=CP (CPCT) and ∠APB=∠APC=90° (linear pair, equal angles) → AP⊥bisector of BC. ✅

Q2. AD is altitude of isosceles △ABC with AB=AC. Show D is midpoint of BC.
Answer: In △ABD and △ACD: AB=AC, AD=AD (common), ∠ADB=∠ADC=90°. ∴ △ABD≅△ACD by RHS. ∴ BD=DC by CPCT. So D is midpoint of BC. ✅

Q3. Two sides AB and BC and median AM of △ABC are respectively equal to PQ, QR and median PN of △PQR. Show △ABC≅△PQR.
Answer: AM is median → BM=½BC. PN is median → QN=½QR. BC=QR → BM=QN. In △ABM and △PQN: AB=PQ, BM=QN, AM=PN. ∴ △ABM≅△PQN by SSS. ∴ ∠ABM=∠PQN → ∠ABC=∠PQR. In △ABC and △PQR: AB=PQ, BC=QR, ∠ABC=∠PQR. ∴ △ABC≅△PQR by SAS. ✅

Q4. BE and CF are two equal altitudes of △ABC. Using RHS rule, prove △ABE≅△ACF. Then prove AB=AC.
Answer: In △ABE and △ACF: BE=CF (given), BC=BC... wait: ∠AEB=∠AFC=90°, hypotenuse AB=AC? No. In △BEC and △CFB: BE=CF, BC=BC, ∠BEC=∠CFB=90°. ∴ △BEC≅△CFB by RHS. ∴ ∠BCE=∠CBF → ∠ABC=∠ACB. In △ABC, ∠B=∠C → AB=AC (sides opposite equal angles). ✅

Q5. ABC is isosceles with AB=AC. Draw AP⊥BC. Show ∠B=∠C.
Answer: In △ABP and △ACP: AB=AC (given), AP=AP (common), ∠APB=∠APC=90°. ∴ △ABP≅△ACP by RHS. ∴ ∠ABP=∠ACP → ∠B=∠C. ✅