Suppose that $P$ and $Q$ are points on the sides $AB$ and $AC$ respectively of $△ABC$. The perpendiculars to the sides $AB$ and $AC$ at $P$ and $Q$ respectively meet at $D$, an interior point of $△ABC$. If $M$ is the midpoint of $BC$, prove that $PM = QM$ if and only if $∠BDP=∠CDQ$.
What i don't get is that how can i relate the sides with the angles. I tried to extend some segments and find similar triangles, but to no avail.
First of all continue segments $CD,BD$ to intersect with $AB,AC$ in two points. Then there will be a cyclic quadrilateral and the problem will be something like the figure and we have to prove that $NF=NE$
For proving this we will name midpoints of $DG,CG$ as $T,S$ respectively. Then we'll prove that $ETN,FSN$ are equal triangles:
$i) SN=GD/2=ET$
$ii) TN=GC/2=FS$
$ETN=ETG+NTG=180-2×EGD+180-CGD=360-2×EGD-CGD$
$FSN=FSG+NSG=180-2×CGF+180-CGD=360-2×CGF-CGD$
And because $CGF=DGE$,
$iii) ETN=FSN$
So the two triangles are equal and thus $NF=NE$.
