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March 19, 2017

Two similar geometry problems based on perpendiculars to cevians

Filed under: mathematics — ckrao @ 7:18 am

In this post I wanted to show a couple of similar problems that can be proved using some ideas from projective geometry.

The first problem I found via the Romantics of Geometry Facebook group: let $M$ be the point of tangency of the incircle of $\triangle ABC$ with $BC$ and let $E$ be the foot of the perpendicular from the incentre $X$ of the $\triangle ABC$ to $AM$. Then show $EM$ bisects $\angle BEC$.

The second problem is motivated by the above and problem 2 of the 2008 USAMO: this time let $AM$ be a symmedian of $ABC$ and $E$ be the foot of the perpendicular from the circumcentre $X$ of $\triangle ABC$ to $AM$. Then show that $EM$ bisects $\angle BEC$.

Here is a solution to the first problem inspired bythat of Vaggelis Stamatiadis. Let the line through the other two points of tangency $P, Q$ of the incircle with $ABC$ intersect line $BC$ at the point $N$ as shown below. Note that since $AP$ and $AQ$ are tangents to the circle, line $NPQ$ is the polar of $A$ with respect to the incircle.

Since $N$ is on the polar of $A$, by La Hire’s theorem, $A$ is on the polar of $N$. The polar of $N$ also passes through $M$ (as $NM$ is a tangent to the circle at $M$). We conclude that the polar of $N$ is the line through $A$ and $M$.

Next, let $MN$ intersect $PQ$ at $R$. By theorem 5(a) at this link, the points $(N, R, P, Q)$ form a harmonic range. Since the cross ratio of collinear points does not change under central projection,  considering the projection from $A$, $(N,M,B,C)$ also form a harmonic range. (Alternatively, this follows from the theorems of Ceva and Menelaus using the Cevians intersecting at the Gergonne point and transveral $NPQ$). Also, $NE \perp EM$ as both $NI$ and $IE$ are perpendicular to polar $AM$ of $N$.

Considering a central projection from $E$ of line $NMBC$ to a line $N', M, P', C'$ parallel to $NE$ through $M$, we see that $(N', M, P', C')$ form a harmonic range. Since $N'$ is a point at infinity, this implies $M$ is the midpoint of $B'C'$ and so triangles $B'EM$ and $C'EM$ are congruent (equality of two pairs of sides and included angle is $90^{\circ}$). Hence $EM$ bisects $\angle BEC$ as was to be shown.

For the second problem, we use the following characterisation of a symmedian: $AM$ extended concurs with the lines of tangency of the circumcircle at $B$ and $C$. (For three proofs of this see here.)

Define $N$ as the intersection of $XE$ with $BC$ and $D$ as the intersection of $AM$ with the tangents at $B, C$. Note that line $NBMC$ is the polar of $D$ with respect to the circumcircle. By La Hire’s theorem, $D$ must be on the polar of $N$. This polar is perpendicular to $NX$ (the line joining $N$ to the centre of the circle) and as $ED \perp EX$ by construction of $E$, it follows that line $AEMD$ is the polar of $N$. Again by theorem 5(a) in reference (2), $(N, M, B, C)$ form a harmonic range. Following the same argument as the previous proof, this together with $NE \perp EM$ imply $EM$ bisects $\angle BEC$ as required.

By similar arguments, one can prove the following, left to the interested reader. If $X$ is the $A$-excentre of $\triangle ABC$, $M$ the ex-circle’s point of tangency of $BC$, and $E$ the foot of the perpendicular from $X$ to line $AM$, then $EM$ bisects $\angle BEC$.

References

(1) Alexander Bogomolny, Poles and Polars from Interactive Mathematics Miscellany and Puzzles http://www.cut-the-knot.org/Curriculum/Geometry/PolePolar.shtml, Accessed 19 March 2017

(3) Yufei Zhao, Lemmas in Euclidean Geometry