Question

ABC is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to equal sides $AC$ and $AB$ respectively. Show that $\Delta ABE\cong \Delta ACF$

Answer 1

In $△ABE$ and $△ACF$, we have

$\angle AEB=\angle AFC$ [Since Each $={90}^o$]

$\angle BAE=\angle CAF$ [Common]

and $AB=AC$ [Given]

$\therefore$ By AAS criterion of congruence, we have

$△ABE\cong △ACF$