Question 168
$ABC$ is an acute triangle. $AE$ and $AF$ are isogonal cevians, where $E \in BC$ and $F \in BC$. The lines $AE$ and $AF$ intersect the circumcircle of $ABC$ again at points $M$ and $N$, respectively. On the rays $AB$ and $AC$ we take points $P$ and $R$ such that $\angle PEA = \angle B$ and $\angle AER = \angle C$. Let $L = AE \cap PR$ and $D = BC \cap LN$. Prove, with reasons, that
$$
\frac{1}{MN} + \frac{1}{EF} = \frac{1}{ED}.
$$
$$
\frac{1}{MN} + \frac{1}{EF} = \frac{1}{ED}.
$$