Question

In a $\Delta ABC,AB=AC.$ If the bisectors of $\angle B$ and $\angle C$ meet $AC$ and $AB$ at points $D$ and $E$ respectively, show that :
(i) $\Delta DBC\cong \Delta ECB$
(ii) $BD=CE$

Answer 1

$AB=AC$

EC is bisector of $\angle C$

BD is bisector of $\angle B$

(i) $AB=AC\to \angle B=\angle C$ (isosceler triangle)

in $\Delta DBC$ and $\Delta ECB$

$BC\to commonside$ ..(1)

$\angle DBC=1/2\angle B=1/2\angle C$ $(\because \angle B=\angle C)$

$\angle ECB=1/2\angle B$

$\Rightarrow \angle DBC=\angle ECB$ ...(2)

$\angle EBC=\angle B=\angle C=\angle DBC$

$\Rightarrow \angle EBC=\angle DCB$ ...(3)

From (1), (2) & (3)

$\Delta DBC\cong \Delta ECB$

$\Rightarrow BD=CE$ $(\because \Delta DBC\cong ECB)$