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P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.

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Solution.
Given : ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA.


To prove : PQRS is a rhombus.
Proof : 
To \triangle ADC, S and R are the mid-points of AD and DC respectively.
By mid point theorem we get 
SR\parallel AC   and    SR=\frac{1}{2}AC              
…..(1)
In
\triangle ABC, P and Q are the mid-points of AB and BC respectively.
Then by mid-point theorem we get
PQ\parallel AC    and     PQ=\frac{1}{2}AC
     …..(2)
From equation 1 and 2 we get
SR=PQ=\frac{1}{2}AC                               …..(3)
Similarly in
\triangle BCD we get
RQ\parallel BD    and      RQ=\frac{1}{2}BD         …..(4)
And in
\triangle BAD
SP\parallel BD  and SP=\frac{1}{2}BD             …..(5)
Using equation 4 and 5 we get
RQ=SP=\frac{1}{2}BD
It is given that AC = BD

 RQ=SP=\frac{1}{2}AC          ......(6)
Now equate equation 3 and 6 we get
SR = PQ = SP = RQ

It shows that all sides of a quadrilateral PQRS are equal.
Hence PQRS is a rhombus.
Hence Proved.

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