Question

prove that quadrilateral formed by joinig the mid point of the adjecent sides of a rectangle is a rhombus in the figure below given that EFGH is a rectangle and PQ,R,S ARE MID POINT of side EF,FG,GH and HE respectively

Answer 1

given- a rectangle EFGH with mid points of P,Q,R,S of sides EF,FG,GH,HE respectively.

to prove- EFGH is a rhombus

construction- join HF

proof- in ∆EHF, S and P are mid points of side EH and EF,
so, SP || HF and SP=1/2HF ........(i)

in ∆HGF, R and Q are mid points of side HG and FG,
so, RQ || HF and RQ=1/2HF ........(ii)

on comparing (i) and (ii)
SP || RQ and SP = RQ
so PQRS is a parallelogram.

now, in ∆EPS and PFQ,
EP=PF (P is mid point of EF)
angle SEP=PFQ (both 90°)
ES=FQ (1/2 EH = 1/2 FG)
so these ∆s are congruent by SAS rule.
so SP=PQ (by cpct)

similarly SR=RQ

in a parallelogram opposite sides are equal so in this way all the sides are equal.

so PQRS is a parallelogram with all sides equal
it is a rhombus.