Question

Question 3
ABC is an isosceles triangle with AB = AC and D is a point on BC such that $AD\perp BC$ (see figure). To prove that $\angle BAD=\angle CAD$, a student proceeded as follows.

In $\Delta ABD$ and $\Delta ACD$, we have AB= AC [given]
$\angle B=\angle C$ [because AB=AC]
And $\angle ADB=\angle ADC$
Therefore,$\Delta ABD\cong \Delta ACD$ [by AAS congruence rule]
So,$\angle BAD=\angle CAD$ [by CPCT]
What is the defect in the above arguments?

Answer 1

In $\Delta ABC,AB=AC$
$\Rightarrow \angle ACB=\angle ABC$ [angle oppsoite to the equal sides are equal]
In $\Delta ABD$ and $\Delta ACD$,

AB=AC [given]
$\angle ABD=\angle ACD$ [proved above]
$\angle ADB=\angle ADC$ [each ${90}^{\circ }]$
$\therefore \Delta ABD\cong \Delta ACD$ [by AAS]
So, $\angle BAD=\angle CAD$ [by CPCT]
So, the defect in the given argument is that firstly prove $\angle ABD=\angle ACD$
Hence, $\angle ABD=\angle ACD$ is defect.