Question
In the figure, $AC = AE, AB = AD$ and $\angle BAD = \angle EAC.$ Prove that $BC = DE.$
✓
Answer
In $\triangle ADE$ and $\triangle BAC$
$AE = AC$
$AB = AD$
$\angle BAD = \angle EAC$
$\angle DAC = \angle DAC = DAC ...($common$)$
$\Rightarrow \angle BAC = \angle EAD = EAD$
Therefore, $\triangle ADE ≅ \triangle BAC ...(\text{SAS}$ criteria$)$
Hence, $BC = DE.$
$AE = AC$
$AB = AD$
$\angle BAD = \angle EAC$
$\angle DAC = \angle DAC = DAC ...($common$)$
$\Rightarrow \angle BAC = \angle EAD = EAD$
Therefore, $\triangle ADE ≅ \triangle BAC ...(\text{SAS}$ criteria$)$
Hence, $BC = DE.$
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