Archives

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

Solution: FromΔ ABD and Δ ADC, we note that ∠ADB = ∠ADC According to the given details AB = BC AD = AD [Common... View Article