Question 3
ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD⊥BC (see figure). To prove that ∠BAD=∠CAD,
a student proceeded as follows. In
ΔABD and
ΔACD, we have AB= AC [given]
∠B=∠C [because AB=AC]
And
∠ADB=∠ADC Therefore,
ΔABD≅ΔACD [by AAS congruence rule]
So,
∠BAD=∠CAD [by CPCT]
What is the defect in the above arguments?