Question

In Fig.3, ABCD and PQRC are rectangles and Q is the mid-point of AC. Prove that
(i) DP = PC
(ii) $PR=\frac{1}{2}AC$

Answer 1

i) This is simple application of mi-point theorem.
Now consider triangle CAD:
Q is midpoint of AC and QP is parallel to AD or CB.
Hence, from mid point theorem (converse) QP bisects the other side which is CD.
So, DP = PC proved.
Further mid point theorem also says, that the line joining mid points of two sides is parallel to third side and equal to $\frac{1}{2}$ of third side.
$\to QP=\frac{1}{2}AD$

(ii) From the above, it is clear that the sides of rectangle PQRC are half of rectangle ABCD;
$PC=\frac{1}{2}CD=\frac{1}{2}AB$
$QP=\frac{1}{2}AD=\frac{1}{2}BC$
Therefore, the diagonal of PQRC must be half the length of diagonal of ABCD.
$\to PR=\frac{1}{2}AC$ since, PR is diagonal of PQRC and AC is diagonal of ABCD.