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, 2023

Geometry Difficulty 8.6 Shortlist Prove it Baltic Way

In an acute triangle $\triangle ABC$ with $|AB| \neq |AC|$ , the angle bisector of $\angle BAC$ intersects side $BC$ and $\odot(ABC)$ at the points $D$ and $M_A$ , respectively. Let points $X$ and $Y$ be the feet of perpendiculars from $M_A$ to sides $AB$ and $AC$ , respectively. The tangent of $\odot(BXM_A)$ at the point $X$ and the tangent of $\odot(CYM_A)$ at the point $Y$ intersect at the point $T$ . Suppose that lines $AT$ and $BC$ intersect at the point $S$ . Show that $\odot(TSM_A)$ passes through the midpoint of segment $AD$ .

Solution

Therefore, quadrilateral $M_AXTY$ is cyclic as well. Combining this with $AXM_AY$ being cyclic, gives us that $M_AXTAY$ is cyclic. Moreover, note that
$\angle TAM_A = \angle TYM_A = \angle YCM_A = \angle ACM_A.$
This means that $AT$ is tangent to $\odot(ABC)$ . Also note that $\angle STM_A = 180^\circ - \angle M_ATA = 90^\circ$ .

Now let $Z$ be the midpoint of $AD$ , then $\angle SZM_A = 90^\circ$ , since $\triangle SAD$ is an isosceles triangle. Combining this with $\angle STM_A = \angle SZM_A = 90^\circ$ , we have that $STZM_A$ is a cyclic quadrilateral, as desired.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.