Bariz - Problem

Source: KMO 2021 P1

Tags: geometry, angle bisector, circumcircle

Olympiadium

13.11.2021 16:30

Let $ABC$ be an acute triangle and $D$ be an intersection of the angle bisector of $A$ and side $BC$. Let $\Omega$ be a circle tangent to the circumcircle of triangle $ABC$ and side $BC$ at $A$ and $D$, respectively. $\Omega$ meets the sides $AB, AC$ again at $E, F$, respectively. The perpendicular line to $AD$, passing through $E, F$ meets $\Omega$ again at $G, H$, respectively. Suppose that $AE$ and $GD$ meet at $P$, $EH$ and $GF$ meet at $Q$, and $HD$ and $AF$ meet at $R$. Prove that $\dfrac{\overline{QF}}{\overline{QG}}=\dfrac{\overline{HR}}{\overline{PG}}$.

Same solution as above. We know that $GE \parallel HF$ because both lines are perpendicular to $AD$ from which we can conclude that $\triangle QHF \sim \triangle QEG$. From this we get $\dfrac{\overline{QF}}{\overline{QG}}=\dfrac{\overline{HF}}{\overline{EG}}$. Now notice that $\angle HFR =\angle PEG$ and also $\angle RHF = \angle DAE = \angle PEG$ from which we get $\triangle GEP \sim \triangle HFR$. This directly gives $\dfrac{\overline{QF}}{\overline{QG}}=\dfrac{\overline{HR}}{\overline{PG}}$. Also in this configuration , $P,Q,R$ are collinear by pascal's theorem on $AEHDGF$.(Just noticed it , not relevant to the problem though)