TODAY'S PROBLEM: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.
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.
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Source: blog.cheenta.com
Source: blog.cheenta.com
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