Library / China Girls' Mathematical Olympiad /32 of 51

China Girls' Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it China

As shown in the figure, quadrilateral $ABCD$ is inscribed in a circle with $AC$ as its diameter, $BD \perp AC$ , and $E$ the intersection of $AC$ and $BD$ . Extend line segment $DA$ and $BA$ through $A$ to $F$ and $G$ respectively, such that $DG \parallel BF$ . Extend $GF$ to $H$ such that $CH \perp GH$ . Prove that points $B$ , $E$ , $F$ and $H$ lie on one circle. (posed by Liu Jiangfeng)

Solution

As shown in the figure, connect $BH$ , $EF$ and $CG$ . Since $\triangle BAF \sim \triangle GAD$ , we have
$\frac{FA}{AB} = \frac{DA}{AG}. \qquad \textcircled{1}$
Furthermore, $\triangle ABE \sim \triangle ACD$ , then
$\frac{AB}{EA} = \frac{AC}{DA}. \qquad \textcircled{2}$
Multiplying ① by ②, we get $\frac{FA}{EA} = \frac{AC}{AG}$ , then $\triangle FAE \sim \triangle CAG$
as $\angle FAE = \angle CAG$ , and thus $\angle FEA = \angle CGA$ .
As is known that $\angle CBG = \angle CHG = 90^\circ$ , then points $B$ , $C$ , $G$ and $H$ lie on one circle. So,
$\begin{aligned} \angle BHF + \angle BEF &= \angle BHC + 90^\circ + \angle BEF \\ &= \angle BGC + 90^\circ + \angle BEF \\ &= \angle FEA + 90^\circ + \angle BEF = 180^\circ. \end{aligned}$
That means that points $B$ , $E$ , $F$ and $H$ lie on one circle.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.