Question

In the given figure, $AB=AC$ and $\angle DBC=\angle ECB={90}^{\circ }$, then $AD=AE$ .

• A. True
• B. False

Answer 1

The correct option is A True

In triangle ABC,
$AB=AC$ thus, $\angle ABC=\angle ACB$ (I) (Opposite angles of equal sides of triangle are equal)
Now, in $△$s DBC and ECB,
$BC=BC$ (Common)
$\angle DBC=\angle ECB=90$ (Given)
$\angle EBC=\angle DCB$ (From I)
Hence, $△DBC\cong △ECB$ (ASA Congruency)
or $DC=BE$ (equal sides of triangle of opposite angle are equal)
Also, Given $AB=AC$
$DC-AC=BE-AB$
$AD=AE$