In the figure, ABCD is a square. AB = EB and $\angle BAE = 70^\circ$. Find $\angle BCE$. Ans: ___ $^\circ$

1. Find $\angle ABE$ using the isosceles triangle $ABE$.

Since $AB = EB$, $\triangle ABE$ is an isosceles triangle.
$\angle BEA = \angle BAE = 70^\circ$
$\angle ABE = 180^\circ - 70^\circ - 70^\circ = 40^\circ$

2. Find $\angle EBC$ using the property of the square.

Since $ABCD$ is a square, $\angle ABC = 90^\circ$.
$\angle EBC = 90^\circ - 40^\circ = 50^\circ$

3. Find $\angle BCE$ using the isosceles triangle $BCE$.

Since $ABCD$ is a square, $AB = BC$. Given $AB = EB$, we have $BC = EB$.
Therefore, $\triangle BCE$ is an isosceles triangle.
$\angle BCE = \frac{180^\circ - 50^\circ}{2} = \frac{130^\circ}{2} = 65^\circ$