Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National Olympiad Prove it JBMO

Problem:

Let $\triangle ABC$ be an acute triangle. The lines $(\varepsilon_{1}),(\varepsilon_{2})$ are perpendicular to $AB$ at the points $A$ , $B$ , respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the sides of the triangle $AC; BC$ intersect the lines $(\varepsilon_{1}),(\varepsilon_{2})$ at the points $E, F$ , respectively. If $I$ is the intersection point of $EF, MC$ , prove that
$\angle AIB = \angle EMF = \angle CAB + \angle CBA$

Solution

Solution:

Let $H, G$ be the points of intersection of $ME, MF$ , with $AC, BC$ respectively. From the similarity of triangles $\triangle MHA$ and $\triangle MAE$ we get
$\frac{MH}{MA} = \frac{MA}{ME}$
thus, $MA^{2} = MH \cdot ME$

Similarly, from the similarity of triangles $\triangle MBG$ and $\triangle MFB$ we get
$\frac{MB}{MF} = \frac{MG}{MB}$
thus, $MB^{2} = MF \cdot MG$

Since $MA = MB$ , from (1), (2), we have that the points $E, H, G, F$ are concyclic.

Therefore, we get that $\angle FEH = \angle FEM = \angle HGM$ . Also, the quadrilateral $CHMG$ is cyclic, so $\angle CMH = \angle HGC$ . We have
$\angle FEH + \angle CMH = \angle HGM + \angle HGC = 90^{\circ}$
Thus $CM \perp EF$ . Now, from the cyclic quadrilaterals $FIMB$ and $EIMA$ , we get that $\angle IFM = \angle IBM$ and $\angle IEM = \angle IAM$ . Therefore, the triangles $\triangle EMF$ and $\triangle AIB$ are similar, so $\angle AIB = \angle EMF$ . Finally
$\angle AIB = \angle AIM + \angle MIB = \angle AEM + \angle MFB = \angle CAB + \angle CBA$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.