Question

Please solve it fast​

Answer 1

Solution

In ∆ ADC `Q` is the midpoint of AC such that PQ || AD &

Converse of mid-point theorem states that in a triangle line drawn from the mid-point of the one side of triangle, parallel to the other side intersect the third side at its mid-point.

So, by theorem we can say DP = PC

In ∆ PQR & ∆ CQR

PQ = CR .....(Side of rectangle PQRC)

QR = QR .....(Same base)

$\angle{PQR} \: = \: \angle{CRQ}$ .....(90° angle)

PR = QC ...{by C.P.C.T}

Now,

2QC = AC ....{midpoint of AC}

2PR = AC

$PR \: = \: \dfrac{1}{2}AC$

Hence Proved