Beautiful geometry Prove $\overline{MN} \parallel \overline{PQ}$.

Line $$\ell$$ intersects sides $$\overline{AB},\overline{AC}$$ of $$\triangle ABC$$ at $$D,E$$. $$P,Q$$ are the midpoints of $$\overline{CD}$$ and $$\overline{BE}$$ respectively. The lines through $$P,Q$$ perpendicular to $$\ell$$ meet the perpendicular bisectors of $$\overline{AC}$$ and $$\overline{AB}$$ at $$M,N$$ respectively. Prove that $$\overline{MN} \parallel \overline{PQ}$$.

The way I approached it was that $$P, Q$$ lie on medial triangle of $$ABC$$ and circumcenter $$O$$ of (ABC) is orhocenter of medial triangle So by taking homothety at $$G$$(centroid of $$ABC$$) with factor of $$2$$ maps medial triangle to $$ABC$$ and $$O$$ to orthocenter of ABC . Let ,$$CX$$ and $$BY$$ be perps on $$AB$$ and $$AC$$ and $$PM$$ intersect $$BY$$ at $$R$$ and $$QN$$ intersect $$CX$$ at $$S$$ then $$RS$$ is parallel to $$MN?$$

I checked it with geogebra but its not true could you point out the mistake please

• Let M1 be midpt of AB then M1Q is parallel to AC and a line parallel to AC through M1 will pass through midpt of BC
– Ken
Dec 30, 2020 at 13:30
• Please see triangle ABE then M1Q is parallel to AE i e AC as M1 is midpt of AB and Q of BE then by thales theorem
– Ken
Dec 30, 2020 at 13:39
• However, I still think that a picture from Geogebra would be helpful; especially, since your approach is difficult to follow (at least for me 😅). For instance, when you refer to $PM$ at the end, do you mean the actual $P,M$, or their images after the homothety. Also: do you want the solution to the problem, or rather to know where your approach fails? Dec 30, 2020 at 14:00
• Yes sir you are very right by PM i mean actual PM and for sone reason i am not able to add a picture very sorry for that, and I want to know both sol and the failiure 😅😅 also if possible could you tell how to correct it??
– Ken
Dec 30, 2020 at 14:08
• The image of $M$ under the homothety is the intersection of $BY$ with the image of $PM$ under the homothety, not the line $PM$ itself. The points $R,S$ have no reason to be parallel to $MN$. Dec 30, 2020 at 14:14

Since $$QN\parallel PM\perp \ell$$, it is enough to show that $$QN=PM$$.
$$P$$ and $$Q$$ lie on the sides of the medial triangle of $$ABC$$.
Let $$M_2$$ be the symmetric of $$C$$ wrt $$M$$, let $$N_2$$ be the symmetric of $$B$$ wrt $$N$$ and let us focus on the pentagon $$AM_2 DE N_2$$: we know that $$M_2 D\parallel N_2 E\perp DE$$ and $$AN_2\perp AD, AM_2\perp AE$$. Both $$AEDM_2$$ and $$AN_2 ED$$ are cyclic quadrilaterals since they have opposite right angles. This means that both $$M_2$$ and $$N_2$$ lie on the circumcircle of $$ADE$$. Since $$\angle M_2 AD = \angle EAN_2$$ it follows that $$M_2 D = N_2 E$$ and $$NQ=MP$$ as wanted.