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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 such that AC \perp BD. Prove that PQRS is a rectangle.

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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 and AC \perp BD.                                       To prove: PQRS is a rectangle
It is given that AC \perp BD

\
\angle COD =\angle AOD = \angle AOB = \angle COB = 90^{\circ}

In \triangle ADC, S and R are the mid-points of AD and DC so by mid-point theorem.
SR \parallel AC   and   SR=\frac{1}{2}AC        
 …..(1)
Similarly in
\triangle ABC,
PQ \parallel AC   &   PQ =\frac{1}{2}AC           …..(2)
Using 1 and 2
SR=PQ=\frac{1}{2}AC                        …..(3)  
Similarly, SP\parallel RQ   &   SP=RQ=\frac{1}{2}BD      ……(4)
In quadrilateral EOFR,
OE \parallel FR, OF \parallel ER
\therefore \angle EOF = \angle ERF = 90^{\circ}
{\angle COD = 90^{\circ} because AC \perp BD and opposite angles of a parallelogram are equal}
In quadrilateral RSPQ
\angle SRQ = \angle ERF = \angle SPQ = 90^{\circ}
   {Opposite angles in a parallelogram are equal}
\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}
90^{\circ} + 90^{\circ} + \angle RSP + \angle RQP = 360^{\circ}
\angle RSP + \angle RQP = 180^{\circ}
\angle RSP =\angle RQP = 90^{\circ}

If all the angles in parallelogram are 90^{\circ} then that parallelogram is a rectangle.  
So, PQRS is a rectangle.
Hence Proved

 

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