In an ABC triangle. plot the height AH, then $ HM \perp AB$ and $HN \perp AC$. Calculate $MN$. if the perimeter of the pedal triangle (DEH) of the triangle ABC is 26 (Answer:13)

My progress: I made the drawing and I believe that the solution must lie in the parallelism and relationships of an cyclic quadrilateral enter image description here


If we reflect $H$ across $AB$ and $AC$ we get two new points $F$ and $G$.

enter image description here

Since $BE$ and $CD$ are angle bisector for $\angle DEH$ and $\angle HDE$ we see $D,E,F$ and $G$ are collinear. Now $MN$ is midle line in the triangle $HGF$ with respect to $FG$ which lenght is \begin{align}FG &= FD+DE+EG\\ &= DH+DE +EH\\&=26 \end{align} so $$ MN = {1\over 2}FG = 13$$

  • $\begingroup$ excellent..thank you for the help $\endgroup$ Sep 23 at 20:03
  • $\begingroup$ would not be MN is midle line in the triangle HGF? $\endgroup$ Sep 23 at 20:23
  • $\begingroup$ "Since BE and CD are angle bisector"...How did you reach this conclusion? $\endgroup$ Sep 23 at 20:28
  • $\begingroup$ Last one is pretty known property of the pedal triangle with respect to $H$. Try to google it. $\endgroup$
    – Aqua
    Sep 23 at 20:33
  • $\begingroup$ Did not know this property ... thank you $\endgroup$ Sep 23 at 21:28

If you know that the orthocenter of the parent triangle is the incenter of the pedal triangle then the work can be made easier. Otherwise as you mentioned, we can always show it using the inscribed angle theorem and the midpoint theorem but it is not as quick as the other answer.

enter image description here

I will refer to the angles of $\triangle ABC$ as $\angle A, \angle B$ and $\angle C$.

We see quadrilateral $BDOH$ is cyclic.

$\angle OHD = \angle OBD = 90^\circ - \angle A$

$\angle DHM = 90^\circ - \angle OHD - \angle BHM$

$ = 90^\circ - (90^\circ - \angle A) - (90^\circ - \angle B) = \angle A + \angle B - 90^\circ$

$ = 180^0 - \angle C - 90^\circ = 90^\circ - \angle C$

Also given $AMHN$ is cyclic,

$\angle HMN = \angle HAN = 90^\circ - \angle C$

In right triangle $ \triangle DMH$, $\angle HMN = \angle DHM$ so $P$ must be circumcenter of the triangle.

Similarly, I will leave it for you to show that $Q$ is the circumcenter of $\triangle ENH$.

Once you show that, $P$ and $Q$ are midpoints of $DH$ and $EH$ respectively, it follows that

$PQ = \frac{DE}{2}, MP = \frac{DH}{2}, NQ = \frac{EH}{2}$

Adding them, $MN = 13$

  • 1
    $\begingroup$ actually with the ownership of incenter it's much easier,,,great explanation $\endgroup$ Sep 23 at 21:30

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