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.
Q.E.D.
In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]
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