Question

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that (i) △ ABE ≅ △ ACF (ii) AB = AC, i.e., ABC is an isosceles triangle.

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?