Let $ABCD$ be a convex quadrilateral. The common external tangents to circles $(ABC)$ and $(ACD)$ meet at point $E$, the common external tangents to circles $(ABD)$ and $(BCD)$ meet at point $F$. Let $F$ lie on $AC$, prove that $E$ lies on $BD$.
Problem
Source: Sharygin 2022 Finals 10-11.2
Tags: geometry
07.08.2022 15:01
Consider the inversion centered at $F$ with radius $r=FB=FD$ mapping $(ABD)$ to $(BCD)$. So, the inversion fixes $B$ and $D$ while $A$ goes to $C$ under this inversion. So, $A'B'=AB\cdot \frac{r^2}{FA\cdot FB}\implies BC=AB\cdot \frac{r^2}{FA\cdot FB}$ and $A'D'=AD\cdot \frac{r^2}{FA\cdot FD}\implies CD=AD\cdot \frac{r^2}{FA\cdot FD}$. Thus, $$AD\cdot BC=AB\cdot CD$$ Now, consider the inversion centered at $E$ with radius $r'=EA=EC$ mapping $(ABC)$ to $(ACD)$. This inversion fixes $A$ and $C$ while $B$ goes to a point $D'\in EB$. Finding the distances between inverted points, we get that $$AD'\cdot BC=AB\cdot CD'$$ Hence, $D'\equiv D$.
28.10.2022 12:03
Denote $O_1,O_2$ to be the orthocenter of $\triangle ABC$ and $\triangle ACD$,and then we can know that: $E$ lies on $BD \Leftrightarrow \angle O_1BD = \angle O_2DB \Leftrightarrow \angle BDC - \angle DBC = \angle BAC = \angle DAC \cdots (*)$ Similarly we have:$F$ lies on $AC \Leftrightarrow \angle ADB - \angle CDB = \angle ACB - \angle CAB \cdots (**)$ Now subtract $\angle DBC + \angle BCA$ on the left and right side of $(**)$ we can get $(*)$.Done!