Bariz - Problem

Source: 2017 Saudi Arabia JBMO TST 2.3

Tags: geometry, Fixed point, perpendicular, Tangents, orthocenter

parmenides51

28.05.2020 07:29

Let $(O)$ be a circle, and $BC$ be a chord of $(O)$ such that $BC$ is not a diameter. Let $A$ be a point on the larger arc $BC$ of $(O)$, and let $E, F$ be the feet of the perpendiculars from $B$ and $C$ to $AC$ and $AB$, respectively. 1. Prove that the tangents to $(AEF)$ at $E$ and $F$ intersect at a fixed point $M$ when $A$ moves on the larger arc $BC$ of $(O)$. 2. Let $T$ be the intersection of $EF$ and $BC$, and let $H$ be the orthocenter of $ABC$. Prove that $TH$ is perpendicular to $AM$.

Given the circle, midpoint $O(0,0)$ and radius $r$. Points $B(b,\sqrt{r^{2}-b^{2}}),C(b,-\sqrt{r^{2}-b^{2}})$ and $D(b,0)$. $BE \bot AC \Rightarrow \triangle BEC$ is right. $BF \bot FC \Rightarrow \triangle BFC$ is right. Then $EBFC$ is concyclic, midpoint of the circle is $D$ and $DE=DF$. $\triangle AEF\ :$ perpendicular bisector of $AE\ //\ BE$ and perpendicular bisector of $EF\ //\ FC$, both lines cut in the point $G \Rightarrow EG=GF$. $EGFD$ is a rhombus. The tangent lines always cut in the fixed point $D$.