Question 36

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Fig. 7.13). To prove that ∠BAD = ∠CAD, a student proceeded as follows: In ∆ ABD and ∆ ACD, AB = AC (Given) ∠B = ∠C (because AB = AC) and ∠ADB = ∠ADC Therefore, ∆ ABD ≅ ∆ ACD (AAS) So, ∠BAD = ∠CAD (CPCT) What is the defect in the above arguments? Fig. 7.13 [Hint: Recall how ∠B = ∠C is proved when AB = AC].

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