Let $ABC$ be a triangle in which (${BL}$is the angle bisector of ${\angle{ABC}}$ $\left( L\in AC \right)$, ${AH}$ is an altitude of$\vartriangle ABC$ $\left( H\in BC \right)$ and ${M}$is the midpoint of the side ${AB}$. It is known that the midpoints of the segments ${BL}$ and ${MH}$ coincides. Determine the internal angles of triangle $\vartriangle ABC$.
Problem
Source: 2011 JBMO Shortlist G3
Tags: geometry, JBMO
Marwanshtine
28.02.2019 12:07
$ABC$ is equilateral triangle
DeZade2002
28.02.2019 12:58
parmenides51 wrote: Let $ABC$ be a triangle in which (${BL}$is the angle bisector of ${\angle{ABC}}$ $\left( L\in AC \right)$, ${AH}$ is an altitude of$\vartriangle ABC$ $\left( H\in BC \right)$ and ${M}$is the midpoint of the side ${AB}$. It is known that the midpoints of the segments ${BL}$ and ${MH}$ coincides. Determine the internal angles of triangle $\vartriangle ABC$. Do you mean that $BL$ and $MH$ intersects each other at their midpoints?
Draw the segment $\overline{ML}$. Let $\overline{BL} \cap \overline{MH} = X$. Now $\triangle BXM$ and $\triangle LXM$ are congruent, so $\angle XLM = \angle XBM = \angle XBH$. So, $ML \parallel BH \implies L$ is the midpoint of $AC \implies AB = BC$.
Again, $\triangle BXM$ and $\triangle BXH$ are congruent, so $BM = BH \implies AM = CH \implies BH = CH \implies H$ is the midpoint of $BC \implies AB = AC$.
So $AB = BC = AC$ making all the internal angles of $\triangle ABC$ $\boxed{60^{\circ}}$.
parmenides51
10.08.2019 00:42
my hint solution