Given in quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.
Also, AC=BD and AC⊥BD.
To prove PQRS is a square.
Proof Now, in ΔADC, S are R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,
SR∥AC and PQ=12AC ……(i)
In
ΔABC,P and Q are the mid-points of AB and BC , then by mid- point theorem,
PQ||AC and PQ=12AC …(ii)
From Eqs. (i) and (ii) ,
PQ||SR and PQ=12AC …(iii)
Similarly, in
ΔABD, by mid-point theorem,
SP||BD and SP=12BD=12AC [given, AC = BD] .....(iv)
And
ΔBCD, by mid-point theorem,
RQ||BD and RQ=12BD=12AC [given, BD = AC] ...(v)
From Eqs. (iv) and (v),
SP=PQ=12AC From Eqs. (iii) and (iv)
PQ = SR = SP= RQ
Thus, all four sides are equal,
Now, in quadrilateral OERF, OE || FR and OF || ER
∴∠EOF∠ERF=90∘ [∵AC⊥DB⇒=∠DOC=∠EOF=90∘ as opposite angles of a parallelogram]
∴∠QRS=90∘ Similarly
∠RQS=90∘ So, PQRS is a square.
Hence proved.