how do I find an angle of an isosceles triangle when it is next to a parallelogram?

Question

$ABCD$ is a parallelogram. $\triangle CDE$ is an isosceles triangle. $\angle DAB \text{ is } 64^°$. Work out the size of the $\angle DEC$.

Answer 1

By the proprieties of parallelogram, we have: $$\angle BAD=\angle CDE$$ Now because $CDE$ is isoscele, we have: $$\angle CED=180°-2\cdot \angle CDE = 180°-2\cdot64°=52°°$$