Let ABC be a triangle such that AC is its shortest side. A point P is inside it and satisfies that BP=AC. Let R be the midpoint of BC and let M be the midpoint of AP. Let E be the intersection of BP and AC. Prove that the bisector of angle ∠BEA is perpendicular to segment MR.
Problem
Source: Mathematics Regional Olympiad of Mexico West 2021 P5
Tags: perpendicular, geometry
10.09.2022 00:16
As M is the midpoint of AP and R is the midpoint of BC, we have the vector equation →RM=→CA+→BP2. Now look at triangle BEC. Since vectors →CA and →BP have the same magnitude, and their directions are along the sides of the triangle, their vector sum must be parallel to the internal angle bisector of ∠BEC. This means →RM is perpendicular to the external angle bisector of ∠BEC, which is ∠BEA, as desired.
11.09.2022 14:24
@ above - that's clever! My way: - Let RM intersect BE at Y and CA at Z. it is enough to prove that EY=EZ. To do so is just two applications of menelaeus, taking YM as the traversal of triangle APE and then RY as the traversal of triangle BEC. It made me laugh how it all suddenly drops out.
30.07.2024 05:42
I don’t see how you can conclude with that so directly. Could you explain more?
31.07.2024 00:36
Let Q be the reflection of P with respect to R. Then BPCQ is a parallelogram, so ∠BEA=∠ACQ, and MR∥AQ is a midsegment in triangle APQ, so it remains to show that the bisector of ∠ACQ is perpendicular to AQ. But this holds since AC=BP=CQ, done.
31.07.2024 02:08
Reflect P over R to get P′. First of all by homothety MR∥AP′. Further BP∥CP′ as BPCP′ is a parallelogram. Hence the bisector of ∠AEB is parallel to the bisector of ∠ACP′. Yet AC=BP=CP′ hence △ACP′ is isoscles, and it is well known the angle bisector is perpendicular to the opposite side.
29.09.2024 01:48
Actually, this is isomorphic to the following well known result (just take P to be outside of ABC to see is properly): if quadrilateral ABCD is such that AB=CD, then with E=AB∩CD the line through the midpoints of AD and BC is parallel to the angle bisector of ∠BEC.