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

  • $\begingroup$ 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 $\endgroup$
    – Ken
    Dec 30, 2020 at 13:30
  • $\begingroup$ 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 $\endgroup$
    – Ken
    Dec 30, 2020 at 13:39
  • 3
    $\begingroup$ 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? $\endgroup$
    – Dr. Mathva
    Dec 30, 2020 at 14:00
  • $\begingroup$ 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?? $\endgroup$
    – Ken
    Dec 30, 2020 at 14:08
  • 1
    $\begingroup$ 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 $MN$. $\endgroup$
    – jlammy
    Dec 30, 2020 at 14:14

1 Answer 1


enter image description here

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.

  • 4
    $\begingroup$ Much better with colors. $\endgroup$ Dec 30, 2020 at 14:58
  • 3
    $\begingroup$ Beautiful solution sir $\endgroup$
    – Ken
    Dec 30, 2020 at 15:19

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