While solving the exercises of my book I came across this interesting problem:
$\triangle ABC$ is isosceles triangle with $AB=AC$. D is a point on base BC such that $AD$ perpendicular on $BC$. To prove that $\angle BAD=\angle CAD$ a student does as follows. Between $\triangle ABD$ and $\triangle ACD$,
- $AB=AC$ (given)
- $\angle B=\angle C$ (because $AB=AC$)
- $\angle ADB=\angle ADC$ ($=90^\circ$).
Therefore $\triangle ABD\cong \triangle ACD$. So, $\angle BAD=\angle CAD$. Now what is the defect in these arguments?
Now what is the defect in this argument? Please try to solve it.