In the diagram above, AB//CD, the bisector of ?BAC and ?ACD meet at E. Find the value of ?AEC

Answer Details

Since AB is parallel to CD, then we have: \begin{align*} \angle AEC &= \angle AED + \angle DEC \\ &= \frac{1}{2} \angle BAC + \frac{1}{2} \angle ACD \\ &= \frac{1}{2}(\angle BAC + \angle ACD) \\ &= \frac{1}{2}(180^\circ) \qquad \text{(because } \angle BAC + \angle ACD = 180^\circ)\\ &= 90^\circ \end{align*} Therefore, the value of $\angle AEC$ is $\boxed{90^\circ}$.