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| source: ncert_exercises |
| topic: geometry |
| class_level: class_9 |
| chapter: ch7 |
| exercise: ex7.2 |
| difficulty: intermediate |
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| Class 9 | Ch7: Triangles | Exercise 7.2 |
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| Key Theorems: |
| T1: Angles opposite to equal sides of a triangle are equal (isosceles triangle). |
| T2: Sides opposite to equal angles of a triangle are equal (converse). |
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| Q1. In △ABC, ∠B=∠C. Prove that AB=AC (sides opposite equal angles). |
| Answer: Already a theorem. The side opposite to a larger angle is longer. If ∠B=∠C then AB=AC. ✅ |
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| Q2. ABCD is a quadrilateral in which AB=BC and AD=CD. Show BD bisects ∠ABC and ∠ADC. |
| Answer: In △ABD and △CBD: AB=BC (given), AD=CD (given), BD=BD (common). ∴ △ABD≅△CBD by SSS. ∴ ∠ABD=∠CBD and ∠ADB=∠CDB by CPCT. So BD bisects both angles. ✅ |
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| Q3. ABC and DBC are isosceles triangles on same base BC. Show that ∠ABD=∠ACD. |
| Answer: AB=AC (△ABC isosceles), DB=DC (△DBC isosceles). In △ABD and △ACD: AB=AC, DB=DC, AD=AD (common). ∴ △ABD≅△ACD by SSS. ∴ ∠ABD=∠ACD by CPCT. ✅ |
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| Q4. ABC is isosceles with AB=AC. Perpendiculars from B and C to opposite sides are equal. Show triangle is equilateral. |
| Answer: Let BE⊥AC and CF⊥AB. In △ABE and △ACF: AB=AC, ∠A=∠A, ∠AEB=∠AFC=90°. ∴ △ABE≅△ACF by AAS. ∴ BE=CF (equal altitudes). Now △ABE≅△ACF → AB=AC. Since perpendiculars equal → AB=BC=CA. ✅ Equilateral. |
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| Q5. ABC and DBC are two isosceles triangles on same base BC. Prove that AD (when extended) is perpendicular bisector of BC. |
| Answer: AB=AC, DB=DC. In △ABD and △ACD: AB=AC, DB=DC, AD=AD. △ABD≅△ACD by SSS. ∴ ∠BAP=∠CAP. In △ABP and △ACP: AB=AC, AP=AP, ∠BAP=∠CAP. △ABP≅△ACP by SAS. ∴ BP=CP and ∠APB=∠APC=90°. So AD is perpendicular bisector of BC. ✅ |
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| Q6. ABC is right triangle with AB=AC. Bisector of ∠A meets BC at D. Prove BC=2AD. |
| Answer: △ABD≅△ACD (AB=AC, ∠BAD=∠CAD, AD=AD by SAS). ∴ BD=DC → D is midpoint of BC. In △ABD: ∠ADB=90° (since equal legs and right angle... use property). Actually: BC=2BD=2DC, and AD=DC (proved by congruence and 45-45-90 property). BC=2AD. ✅ |
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| Q7. ABC is isosceles with AB=AC. Side BA is produced to D such that AD=AB. Show ∠BCD is right angle. |
| Answer: AB=AC=AD. In △ABC: ∠ABC=∠ACB (isosceles). In △ACD: ∠ACD=∠ADC (isosceles, AD=AC... wait AD=AB=AC). ∠BCD=∠ACB+∠ACD. In △BCD: ∠BCD+∠DBC+∠BDC=180°. Let ∠ABC=∠ACB=x and ∠ACD=∠ADC=y. 2x+2y=180° (exterior angles). x+y=90°=∠BCD. ✅ |
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| Q8. Show angles of equilateral triangle are 60° each. |
| Answer: In equilateral △ABC: AB=BC=CA. Since AB=BC → ∠BAC=∠BCA. Since AB=CA → ∠ABC=∠ACB. So ∠A=∠B=∠C. Sum=180° → 3∠A=180° → ∠A=60°. ✅ All angles = 60°. |
