AoPS - Problem

Source: Dutch BxMO TST 2016 p3

Tags: geometry, incircle, collinear, right triangle

parmenides51

03.08.2019 02:49

Let $\vartriangle ABC$ be a right-angled triangle with $\angle A = 90^o$ and circumcircle $\Gamma$. The inscribed circle is tangent to $BC$ in point $D$. Let $E$ be the midpoint of the arc $AB$ of $\Gamma$ not containing $C$ and let $F$ be the midpoint of the arc $AC$ of $\Gamma$ not containing $B$. (a) Prove that $\vartriangle ABC \sim \vartriangle DEF$. (b) Prove that $EF$ goes through the points of tangency of the incircle to $AB$ and $AC$.

Let $I$ be the incenter of $\triangle ABC$ and $G$ the perpendicular from $I$ to $BC$. $(a)$. By angle chasing we have $\angle IEB=\angle CEB=\angle A=90=180-90=180-\angle IDB$ So $IDBE$ is cyclic and $\angle DEF=\angle DEI+\angle CEF=\tfrac{\angle B}{2}+\tfrac{\angle B}{2}=\angle B$ And similarly we show $\angle DFE=\angle C$. $(b)$ Since $\angle IGB=90=\angle IEB$ we have that $IGBE$ is cyclic. So $\angle CEG=\angle IEG=\tfrac{\angle B}{2}=\angle CEF$ So $EGF$ are colinear and similarly we show the other part.