Question
In figure, $A C=A E, A B=A D$ and $\angle B A D=\angle E A C$. Show that $B C=D E$.
✓
Answer
Given : $AC = AE, AB = AD$ and $\angle BAD = \angle EAC.$
To prove ; $BC = DE$
Proof : In $DABC$ and $DADE$
$AC = AE, AB = AD$ and $\angle BAD = \angle EAC ...[$Given]
$\therefore \angle BAD + \angle DAC = \angle DAC + \angle EAC .$..[Adding $\angle DAC $to both sides]
$\therefore \angle BAC = \angle DAE ...(1) $
$ AC = AE ... $ [Given]
$\angle BAC = \angle DAE ...$ [From$ (1)] $
$ AB = AD ... $[Given]
$\therefore DABC \cong DADE ...$ [By $SAS$ property]
$\therefore BC = DE ...$ [c.p.c.t.]
To prove ; $BC = DE$
Proof : In $DABC$ and $DADE$
$AC = AE, AB = AD$ and $\angle BAD = \angle EAC ...[$Given]
$\therefore \angle BAD + \angle DAC = \angle DAC + \angle EAC .$..[Adding $\angle DAC $to both sides]
$\therefore \angle BAC = \angle DAE ...(1) $
$ AC = AE ... $ [Given]
$\angle BAC = \angle DAE ...$ [From$ (1)] $
$ AB = AD ... $[Given]
$\therefore DABC \cong DADE ...$ [By $SAS$ property]
$\therefore BC = DE ...$ [c.p.c.t.]
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