Question

In Fig. AC = AE, AB = AD and $\angle$ BAD = $\angle$ EAC. Prove that BC = DE.
[2 MARKS]

Answer 1

We have,

$\angle$BAD = $\angle$EAC

Then,
$\angle$BAD + $\angle$DAC = $\angle$EAC + $\angle$DAC $\left(0.5Mark\right)$
[Adding $\angle$DAC to both sides]

$\Rightarrow$ $\angle$BAC = $\angle$DAE ---- (i) $\left(0.5Mark\right)$

Now, In $\Delta ABC$ and $\Delta ADE$,
AB = AD [Given]
$\angle$BAC = $\angle$DAE [From (i)]
and, AC = AE [Given]

So,
$\Delta$ABC $\cong$ $\Delta$ADE [SAS congruence criterion] $\left(0.5Mark\right)$

$\Rightarrow$BC = DE [ C.P.C.T ] $\left(0.5Mark\right)$