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Question

# ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that $CQ=\frac{1}{4}AC.$ If PQ produced meets BC at R, prove that R is the midpoint BC.

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Solution

## We know that the diagonals of a parallelogram bisect each other. Therefore, CS = $\frac{1}{2}$AC ...(i) Also, it is given that CQ = $\frac{1}{4}$AC ...(ii) Dividing equation (ii) by (i), we get: $\frac{CQ}{CS}=\frac{\frac{1}{4}AC}{\frac{1}{2}AC}$ or, CQ = $\frac{1}{2}$CS Hence, Q is the midpoint of CS. Therefore, according to midpoint theorem in $△CSD$ $PQ\parallel DS$ $\mathrm{if}PQ\parallel DS,\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}QR\parallel SB$ In $△CSB,Q\mathrm{is}\mathrm{midpoint}\mathrm{of}CS\mathrm{and}QR\parallel SB$. Applying converse of midpoint theorem ,we conclude that R is the midpoint of CB. This completes the proof.

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