In the diagram above, AB//CD, the bisector of ?BAC and ?ACD meet at E. Find the value of ?AEC
Question 1 Report
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 $\boxed{90^\circ}$.
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