Let $\triangle ABC$ be isosceles with $\overline{AB} \cong \overline{AC}$ and altitudes $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ intersecting at $H$, the orthocenter. $\bigcirc G$, of diameter $CE$, intersects $\overline{BC}$ and $\overline{CF}$ at $M$ and $N$, respectively. $\overrightarrow{MN}$ intersects the altitude $\overline{AD}$ at $P$. Prove that $\overline{DP} \cong \overline{ME}$.
So far, I have found that $\angle CME$ is right, because it's an inscribed angle. Geogebra says that $MN=MC$, but I am not sure why. I think the proof might involve the similar $\triangle AFE$, then proving that $FE \parallel BC$, but I really don't know how to get there.
Any help would be really appreciated.
Best Answer
First, note that $\angle MEC = \angle MNC$, since they are determined by the same arc of the circumference. Then, since $\angle PNH$ and $\angle MNC$ are vertically opposite angles, they have the same value and so $\angle MEC = \angle MNC = \angle PNH$.
In quadrilateral $CDHE$, opposite angles $\angle CEH$ and $\angle CDH$ are both equal to $90^\circ$, which means their sum equals $180^\circ$ and so the sum of the other two angles also is $180^\circ$, so $\angle DHE + \angle DCE = 180^\circ$. But $\angle DHE + \angle PHE = 180^\circ$ also, so $\angle DCE = \angle MCE = \angle PHE$.
We have $\angle CNE = 90^\circ$, since $EC$ is the diameter, and so $\angle ENH = 90^\circ$. In the right triangle $\triangle CEM$, angles $\angle MCE = \angle DCE$ and $\angle MEC$ are complementary. Since angle $\angle PNH = \angle MEC$ and angle $\angle PNE$ are complementary, we have $\angle PNE = \angle MCE = \angle PHE$.
In quadrilateral $PENH$, we have $\angle PNE = \angle PHE$, which means the quadrilateral is cyclic and the opposite angles are supplementary, thus since $\angle ENH = 90^\circ$, opposite angle $\angle EPH$ is also a right angle. Then the quadrilateral $EPDM$ has three right angles and is a rectangle, so opposite sides $PD$ and $EM$ have the same length.