# Hard Euclidean Geometry question

Let $I$ be the incenter of a triangle $ABC$ and $M$ be the mid-point of the side $BC$. If the line $IM$ cuts the height $AH$ in the point $E$, show that $AE=r$, where $r$ is the radius of the circle inscribed in $ABC$.

I tried to solve this using "classical" geometry and using analytical geometry and failed. I even drew it in Geogebra as follows: Can someone help?

Thanks

• What about Menelaus theorem? – openspace May 7 '16 at 15:14
• @openspace Using line MIE? How that would help? If you say to use it in line AE, I never did Menelaus like this before. Is it possible? – Gabriel May 7 '16 at 16:04
• Consider the ABMEA – openspace May 7 '16 at 18:09
• But I'm sure that this method is useless – openspace May 7 '16 at 18:17

Let $X$ and $Y$ be points of tangency of incircle and $A$-excircle with side $BC$. It is well-known that $MX=MY$.

Let $XZ$ be diameter of incircle. Consider a homothety with center $A$ that takes incircle to $A$-excircle and deduce that $A,Z,Y$ are collinear. Indeed, let $l$ be tangent to incircle at $Z$. This homothety maps $l$ to a line $m$ tangent to excircle. Of course $l \parallel m$. We see that $m=BC$, because $BC$ is tangent to excircle and parallel to $l$. We conclude that $Z$ is mapped to $Y$, as these are points of tangency of $l,m$ to these circles.

Since $M,I$ are midpoints of $XY, XZ$, it follows that $ZY \parallel IM$. Moreover $AE \parallel ZI$. Therefore $AEIZ$ is a parallelogram. Thus $AE=IZ=r$.

• Cool answer! I'm having a little trouble understanding how the homothety implies that $A,Z,Y$ are collinear. – Gabriel May 7 '16 at 16:01
• I have added some explanation. – timon92 May 7 '16 at 16:50

This is a sketch for a brute-force solution. Let $P$ be the projection of $I$ onto $BC$. Then

1. compute $MP$ and $MH$,
2. use $\dfrac{IP}{EH}= \dfrac{MP}{MH}$ to find $EH$
3. $AE=AH-EH$.
• Tried to compute $MP$ and $MH$ in function of $r$ but failed. You have any tip or idea? – Gabriel May 7 '16 at 16:11