Library / Mongolian National Mathematical Olympiad /4 of 31

Mongolian National Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Mongolia

Let $H$ be the intersection point of the altitudes $AD$ and $BE$ of an acute triangle $ABC$ . The circumcircle of the triangle $ABC$ intersects the circle with diameter $CH$ at the point $K$ other than $C$ . Prove that
$\frac{DK}{KE} = \frac{DH}{HE}.$

Solution

Since $\angle BDH = \angle AEH$ and $\angle BHD = \angle AHE$ , we have $\triangle BHD \sim \triangle AHE$ . Therefore
$\frac{DH}{HE} = \frac{BD}{AE}. \qquad (1)$

Also it is easy to observe that $\angle CDK = \angle CEK$ , and moreover $\angle BOK = \angle AEK$ , $\angle KBD = \angle KAE$ . Hence $\triangle BDK \sim \triangle AEK$ . Therefore
$\frac{BD}{AE} = \frac{DK}{KE}. \qquad (2)$
From (1) and (2), we get
$\frac{DH}{HE} = \frac{BD}{AE} = \frac{DK}{KE}.$

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.