# $\triangle ABC$ and $\triangle ADE$ are isosceles. Show that $\angle BAD=\angle EAC$

I'm not sure if my answer is correct for the following question:

$\triangle ABC$ and $\triangle ADE$ are isosceles. Show that $\angle BAD=\angle EAC$.

My answer to this question is:

• $AB=AD$ (Given, properties of isosceles $\triangle$)
• $AC=AE$ (Given, properties of isosceles $\triangle$)
• $\angle BAD+\angle BAO=\angle EAC+\angle OAC$ ($AO$ bisects $BC$)

Therefore, $\triangle ABC\cong\triangle ADE$ (Side-Angle-Side) and $\angle BAD=\angle EAC$.

Can someone please check if my answer is correct or is there a better way to solve the question?

Your answer is completely wrong; $AB$ need not even equal $AD$. Instead, note that (with $AO\perp BC)$ $$\angle BAO=\angle OAC$$ $$\angle DAO=\angle OAE$$ Subtracting the second equality from the first gives $$\angle BAO-\angle DAO=\angle OAC-\angle OAE$$ $$\angle BAD=\angle EAC$$ as desired.
$AB = AD$ and $AC = AE$ are not correct. Instead,
• Since $AD = AE$, we know that $\angle ADE = \angle AED \implies \angle ADB = \angle AEC$.
• Also, by the equality $AB =AC$, we have $\angle ABD = \angle ACE$.
Using these two, we can directly conclude that $\angle BAD = \angle AEC$ because we know that $\angle ABD + \angle ADB + \angle BAD = \angle ACE+\angle AEC + \angle EAC\ (= 180^\circ)$.