Library / /103 of 462

, 2014

Geometry Difficulty 5.2 AIME, harder Prove it Ireland

A circle is drawn through three vertices $A$ , $B$ , $C$ of a parallelogram $ABCD$ intersecting the side $CD$ internally at $E$ and the side $AD$ internally at $F$ . The line $EF$ meets the line $BC$ at $H$ and the line $BA$ at $K$ .
Prove that the circumcircles of the triangles $CEH$ and $AFK$ both touch the circumcircle of triangle $DEF$ . Prove also that the common tangents at $E$ and $F$ both pass through $B$ .

Solution

Because $DC \parallel AB$ and $BEFA$ is cyclic, we have
$\angle EDF = \angle KAF = 180^\circ - \angle FAB = \angle FEB$ and
$\angle CEB = \angle EBA = 180^\circ - \angle EFA = \angle EFD = \angle EHC,$
the last equality because $AD \parallel BH$ .

The first line shows that BE is tangent to the circle DEF and the second line that BE is tangent to the circle CEH. In a similar way, using that BCEF is cyclic, the remaining statements follow.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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