Let $ABC$ be a triangle, and let $P$ be a point inside it such that $\angle PAC = \angle PBC$. The perpendiculars from $P$ to $BC$ and $CA$ meet these lines at $L$ and $M$, respectively, and $D$ is the midpoint of $AB$. Prove that $DL = DM.$
consider the midpoints of lines $BP$ and $AP$ , $E$ and $F$ respectively. obviously $DF=EL$ and $DE=FM$. A simple angle chasing will give you that $\hat{DFM}=\hat{DEL}$ so: $\triangle DFM \cong \triangle DEL$ and so $DL=DM$.
Just use the fact that $\frac{AM}{PM}=\frac{BL}{PL}$ and use cosine rule in $\triangle{CDL}$ to find out $DL^2$.Now its easy to verify that its equal to $DM^2$