Question

$$ \overline { b } \parallel \overline { c d }, \overline { a d } \parallel \overline { b c }, \overline { a d } \cong \overline { a e } $$ and $$ m \angle a d e = 6 3 ^ { \circ } $$. find $$ m \angle a d c $$.

Explanation:

Step1: Find $\angle AED$

Since $\overline{AD}\cong\overline{AE}$, $\triangle ADE$ is isosceles. So $\angle ADE=\angle AED = 63^{\circ}$.

Step2: Find $\angle DAE$

Using the angle - sum property of a triangle ($\angle ADE+\angle AED+\angle DAE = 180^{\circ}$), we have $\angle DAE=180^{\circ}-63^{\circ}-63^{\circ}=54^{\circ}$.

Step3: Identify the parallelogram

Since $\overline{AB}\parallel\overline{CD}$ and $\overline{AD}\parallel\overline{BC}$, $ABCD$ is a parallelogram.

Step4: Use the property of parallelogram

In parallelogram $ABCD$, $\angle ADC+\angle DAB = 180^{\circ}$ (adjacent angles of a parallelogram are supplementary). And $\angle DAB=\angle DAE = 54^{\circ}$. So $\angle ADC=180^{\circ}-\angle DAB$.

Answer:

$126^{\circ}$