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

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Solution.
Given: ABCD is a parallelogram and P, Q, R and S are the mid-points of sides AB, BC, CD and AD. Also AC = BD and AC \perp BD.
To prove: PQRS is a square

Proof: In \triangle ADC
, S and R are the mid-points of sides AD and DC by mid-point theorem.
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. Therefore, by mid-point theorem
PQ\parallel AC    and    PQ=\frac{1}{2}AC      …..(2)
From equation 1 and 2 we get
PQ\parallel SR     and    PQ=\frac{1}{2}AC     
…..(3)
Similarly, in
\triangle BCD, RQ\parallel BD and R, Q are midpoints of CD, CB respectively, Therefore, by mid-point theorem
RQ=\frac{1}{2}BD=\frac{1}{2}AC         {Given BD = AC} …..(4)
And in \triangle ABD, SP\parallel BD and S, P are midpoints of AD, AB respectively. Therefore, by mid-point theorem

SP=\frac{1}{2}BD=\frac{1}{2}AC    {Given AC = BD  …..(5)
From equation 4 and 5 we get
SP=RQ=\frac{1}{2}AC              …..(6)
From equation 3 and 6 we get

PQ = SR = SP = RQ
Thus, all sides are equal
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}
All the sides are equal and all the interior angles of the quadrilateral are 90^{\circ}

Hence, PQRS is a square.
Hence Proved

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