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:
ΔABD and
ΔACD,AB
= AC (Given)
∠B=∠C (because AB
= AC)
and
∠ADB=∠ADCTherefore,
ΔABD≅ΔACD(AAS)So,
∠BAD=∠CAD(CPCT)What is the defect in the above arguments?