^@ABCD^@ is a parallelogram where ^@P^@ and ^@R^@ are the midpoi | Math Question and Answer

Answer:

Step by Step Explanation:

Let us draw the image for the situation given in the question.
Also, join $BD$ intersecting $AC$ at $O$ .
It is given that $P$ is the mid-point of $BC$ and $R$ is the mid-point of $DC$ .

Thus, in triangle $CBD$ , by using mid-point theorem $PR \parallel BD$ .
As, $\begin{aligned} & PR \parallel BD \\ \implies & PQ \parallel BO \text{ and } QR \parallel OD \end{aligned}$ Now, in triangle $BCO$ , we have $PQ \parallel BO$ and $P$ is the mid point of $BC$ .
By the inverse of mid-point theorem, $Q$ is the midpoint of $OC$ . $\implies 2 CQ = OC$
As the diagonals of a parallelogram bisect each other, $AO = OC$ .
Thus, $\begin{aligned} & AC = AO + OC \\ \implies & AC = 2 OC && [\because \text{ AO = OC]} \\ \implies & AC = 2 \times 2 CQ && [\because \text{ 2 CQ = OC]} \\ \implies & AC = 4 CQ \end{aligned}$

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