INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

September 7, 2012

An application of Menalaus' theorem

Given: AB is the diameter of a circle with center O. C be any point on the circle. OC. is joined. Let Q be the midpoint of OC. AQ produced meet the circle at E. CD be perpendicular to diameter AB. ED and CB are joined.

R.T.P. : CM = MB

Construction: AC and BD joined.

Proof: In triangle BOC, AQF is the transversal. Applying Menalaus' Theorem we have CF/FB = 1/2.
Now ΔABC is similar to ΔBDH and ΔACF is similar to ΔDGH (right angles equal and angle subtended by the same segment equal in both cases).
Hence HG/GB = CF/FB. But CF/FB =1/2. Hence HG/GB = 1/2.
Now applying Menalaus' Theorem in ΔBCH with transversal DGM we have CM/MB=1.


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.