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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.



(1) Alexander Bogomolny, Poles and Polars from Interactive Mathematics Miscellany and Puzzles, Accessed 19 March 2017

(2) Poles and Polars – Another Useful Tool! | The Problem Solver’s Paradise

(3) Yufei Zhao, Lemmas in Euclidean Geometry

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