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.

Answer 1

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:
$\frac{CQ}{CS}=\frac{{\displaystyle \frac{1}{4}}AC}{{\displaystyle \frac{1}{2}}AC}$
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.