Question

In Fig., $D$ and $E$ are points on side $BC$ of a $△ABC$ such that $BD=CE$ and $AD=AE$. Show that $△ABD\cong △ACE$.

Answer 1

$AD=AE$

$\angle ADE=\angle AED$

$\angle ADB+\angle ADE={180}^o$

$\angle ADB=180-\angle ADE$

$BD=CE$

$AD=CE$

$△ABD\cong △ACE$ proved