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In the adjoining figure, AB = AC and BD = DC. Prove that ΔADB ≅ ΔADC and hence show that(i) ∠ADB = ∠ADC = 90o, (ii) ∠BAD = ∠CAD. |
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Answer» Given, AB = AC and BD = DC To prove, ΔADB ≅ ΔADC Proof, In the right triangles ADB and ADC, we have: Hypotenuse AB = Hypotenuse AC (given) BD = DC (given) AD = AD (common) ∴ ΔADB ≅ ΔADC By SSS congruence property: ∠ADB = ∠ADC (corresponding parts of the congruent triangles) … (1) ∠ADB and ∠ADC are on the straight line. ∴∠ADB + ∠ADC =180o ∠ADB + ∠ADB = 180o 2 ∠ADB = 180o ∠ADB = 180/2 ∠ADB = 90o From (1): ∠ADB = ∠ADC = 90o (ii) ∠BAD = ∠CAD (∵ corresponding parts of the congruent triangles) |
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