ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that $\text{CQ}=\frac{1}{4}\text{AC}.$ If PQ produced meets BC at R, prove that R is the midpoint of BC.
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We know that the diagonals of a parallelogram bisect each other.
Therefore,
$\text{CS}=\frac{1}{2}\text{AC}\dots(\text{i})$
Also, it is given that $\text{CQ}=\frac{1}{4}\text{AC}\dots(\text{ii})$
Dividing equation (ii) by (i), we get:
$\frac{\text{CQ}}{\text{CS}}=\frac{\frac{1}{4}\text{AC}}{\frac{1}{2}\text{AC}}$
or, $\text{CQ}=\frac{1}{2}\text{CS}$
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in $\triangle\text{CSD}$
PQ || DS
If PQ || DS, we can say that QR || SB
In $\triangle\text{CSB},\text{Q}$ is midpoint of CS and QR || SB.
Applying converse of midpoint theorem, we conclude that R the midpoint of CB.
This comletes the proof .
Therefore,
$\text{CS}=\frac{1}{2}\text{AC}\dots(\text{i})$
Also, it is given that $\text{CQ}=\frac{1}{4}\text{AC}\dots(\text{ii})$
Dividing equation (ii) by (i), we get:
$\frac{\text{CQ}}{\text{CS}}=\frac{\frac{1}{4}\text{AC}}{\frac{1}{2}\text{AC}}$
or, $\text{CQ}=\frac{1}{2}\text{CS}$
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in $\triangle\text{CSD}$
PQ || DS
If PQ || DS, we can say that QR || SB
In $\triangle\text{CSB},\text{Q}$ is midpoint of CS and QR || SB.
Applying converse of midpoint theorem, we conclude that R the midpoint of CB.
This comletes the proof .
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