Prove that $\overline{DP} \cong \overline{ME}$ in $\triangle ABC$

Question

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 $\overline{DP} \cong \overline{ME}$.

My Diagram

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.

Answer 1

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.

Answer 2

We use the property of nine point circle which passes the mid point of sides and the foot of altitudes, in our case mid point of BC and the foot of altitude AD are coincident.As can be seen in figure EF is the chord of this circle which is parallel with base BC, therefore MP is the diagonal of rectangular PDME. Now angle HEN , where HE is tangent on circle, is opposite the arc EN , so it is equal to angle NCE. But $CE\bot HE$, this means other rays of angles HEN and NCE, i.e NC and NE must also be perpendicular , that is $\angle CNE=90^o$ which mean N is the intersection circle (G,E) as mentioned in statement. Hence $ME=PD$.

Answer 3

According to Andrew's figure:

We have $\angle NME=\angle NCE$ since they see the same arc in the circle.

Since $\triangle FCA$ is right triangle, $\angle NCE=90^\circ-A^\circ$. Hence, have $\angle PMD=\angle NMD=\angle A$.

It is enough to show that $\angle EDC=\angle A.$ This follows from the fact that $\triangle BEC$ is right triangle and D is midpoint of its hypotenus.