Let
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
be a triangle, and let
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
be a point inside it such that
![\angle PAC = \angle PBC](/media/m/f/c/e/fce89dcca8a8d5457e9ae1db125ef12b.png)
. The perpendiculars from
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
to
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
and
![CA](/media/m/a/a/e/aaec86bc003cfdb64d54116a4cabd387.png)
meet these lines at
![L](/media/m/f/c/1/fc1ae4eb78da7d1352cbf1f8217ab286.png)
and
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
, respectively, and
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
is the midpoint of
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
. Prove that
%V0
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 $DL = DM.$