Question

ABCD is a kite where AB =AD and BC =CD .P,Q,R and S are the mid -points of the sides AB, BC, CD and AD respectively. Prove that PQRS is a rectangle.

Answer 1

Given: ABCD is a kite where AB =AD and BC =CD
P,Q,R and S are the mid -points of the sides AB, BC, CD and AD respectively.
To prove : PQRS is a rectangle.
Construction: Join AC and BD, which intersect in O.

Proof: In $\Delta ABD,P$ and S are the mid-points of the AB and AD respectively.
PS||BD and $PS=\frac{1}{2}BD...\left(1\right)$(Mid point theorem)
In $\Delta BCD,Q$ and R are the mid-points of the sides BC and CD respectively.
QR||BD and $QR=\frac{1}{2}BD....\left(2\right)$ (Mid point theorem)
From (1) and (2), PS||QR and PS =QR
$\therefore$ PQRS is a parallelogram ...(3) (In a quadrilateral, if one pair opposite sides is parallel and equal, then it is a parallelogram.)
Since ABCD is a Kite,
$\therefore \angle AOD={90}^{\circ }...\left(4\right)$ (In kite, diagonals intersect each other at ${90}^{\circ })$
Since PS||BD
FS||OE ....(5)
SR||AC (Using mid -point theorem)
$\therefore SE||OF$
Front (5) and (6), we get
OESF is a parallelogram (Both pair of opposite sides are parallelogram)
$\angle FOE=\angle FSE={90}^{\circ }$ (Opposite angles of parallelogram are equal).
$\therefore \angle PSR={90}^{\circ }...\left(7\right)$
From (3) and (7), we can conclude that Parallelogram ABCD is a rectangle.