Length of lines related to orthocenter

Question

Let ABC be a triangle and let H be its orthocenter. Let M be the midpoint of BC. The perpendicular to MH through H intersects AB and AC at P and Q, respectively. Prove that |MP| = |MQ|.

My attempt so far:

Clearly, this can easily be solved using coordinate geometry, but I'm trying to practice my euclidean geometry. One thing I've noticed is that if B' and C' are the feet of C and B to AB and AC respectively, then, in the cyclic quadrilateral BC'B'C with center M, we get that H is the intersection of the quadrilateral's diagonals. I don't know how to proceed from here though.

Limestone · Accepted Answer · 2021-11-03T10:14:33.680

3

Let $AH$, when extended, meet $BC$ at point $D$. Then, $\angle BMH=90-\angle DHM=\angle PHD=\angle AHQ$ and similarly $\angle CMH=\angle AHP$

Observe that, $\triangle BHM\sim \triangle AQH$ since $\angle AHQ=\angle BMH$ and $\angle HAQ=\angle MBH$. $\Rightarrow \frac {BM}{HM}=\frac {AH}{HQ}$

Similarly, $\triangle CHM\sim \triangle APH$ $\Rightarrow \frac {CM}{HM}=\frac {AH}{HP}$.

Comparing these two equalities gives $HP=HQ$ and therefore $MP=MQ$.

edited Nov 03 '21 at 10:14

answered Nov 03 '21 at 10:01

Limestone

2,488

Why are $\angle AHQ=\angle BMH$ and $\angle HAQ=\angle MBH$? – justadumbguy Nov 03 '21 at 10:06
@justadumbguy $AH\perp BC$ and $BH\perp CA$. $\Rightarrow \angle HAQ=90-\angle C$ and $\angle MBH=90-\angle C$. Hence, $\angle HAQ=\angle MBH$. – Limestone Nov 03 '21 at 10:18

score 3 · Answer 2 · answered Nov 03 '21 at 10:20

Let $A'$ be the reflection of $A$ through $H$, and let the parallel to $AC$ through $A'$ intersect $AB$ at $P_1$.

Then the triangles $HBC$ and $P_1 A'A$ have their sides respectively perpendicular. Hence they are homothetic after a quarter turn. It follows that their medians $P_1H$ and $HM$ are perpendicular. Then $P_1 = P$ and the result follows easily from the fact that $APA'Q$ is a parallelogram.

Length of lines related to orthocenter

2 Answers2