Question

in the diagram below, $overline{ad}congoverline{ae}$, $overline{ba}paralleloverline{ce}$, $overline{cb}paralleloverline{da}$ and $mangle bcd = 62^{circ}$. find $mangle e$.

Explanation:

Step1: Identify the parallelogram

Since $\overline{BA}\parallel\overline{CE}$ and $\overline{CB}\parallel\overline{DA}$, quadrilateral $ABCD$ is a parallelogram. In a parallelogram, opposite - angles are equal. So, $\angle BAD=\angle BCD = 62^{\circ}$.

Step2: Consider the isosceles triangle

Given $\overline{AD}\cong\overline{AE}$, $\triangle ADE$ is an isosceles triangle. Let $\angle E=\angle ADE = x$.
The exterior - angle of a triangle is equal to the sum of the two non - adjacent interior angles. The exterior angle of $\triangle ADE$ at $A$ is $\angle BAD$.
We know that $\angle BAD=\angle E+\angle ADE$ (by the exterior - angle property of a triangle).
Since $\angle BAD = 62^{\circ}$ and $\angle E=\angle ADE$, we have $62^{\circ}=x + x$.

Step3: Solve for $\angle E$

Combining like terms in the equation $62^{\circ}=2x$, we get $x=\frac{62^{\circ}}{2}=31^{\circ}$.

Answer:

$m\angle E = 31^{\circ}$