Question

P,q,r and s are midpoints of the sides ab,bc,cd and da respectively of a quadrilateral abcd in which ac=bd. Prove that pqrs is a rhombus

Answer 1

It is given that the points P , Q , R , S are the mid points of AB , BC , CD , DA respectively.


According to the statement :

• AP = BP = AB / 2
• AS = SD = AD / 2
• RD = RC = DC / 2
• BQ = QC = BC / 2



As the given quadrilateral is a rectangle, opposite sides should be equal.
Hence,
• AB = DC

From above,
= > AB / 2 = AP = BP = DC / 2 = RD = RC

= > AP = BP = RD = RC ---: ( 1 )



Also,
• AD = BC

From above,
= > AD / 2 = AS = SD = BC / 2 = BQ = QC

= > AS = SD = BQ = QC ---: ( 2 )



Now, by using Pythagoras Theorem :

In ∆ASP,
• SP² = AP² + AS²

In ∆BPQ,
• QP² = PB² + BQ²

From ( 1 ) & ( 2 ),
= > QP² = ( AP )² + ( AS )²
= > QP² = AP² + AS²
= > QP² = SP²
= > QP = SP


In ∆RDS,
• SR² = SD² + RD²

From ( 1 ) & ( 2 ),
= > SR² = AS² + AP²
= > SR² = SP²
= > SR = SP


In ∆RCQ,
• QR² = QC² + RC²

From ( 1 ) & ( 2 ),
= > QR² = AS² + AP²
= > QR² = SP²
= > QR = SP


Hence,
SR = SP = QR = QP

therefore,
All sides of the formed quadrilateral are equal.



Now, joining the P with D and S with Q,

As AS = BQ = SD = QC, AB || SQ || DC
And, AP = PB = RD = RC, AD || PR || BC


Hence,
Let the intersection of RP & QS be O,

As PB || QO, BQ || OP,
angle PBQ = 90° = angle BPO = angle OPQ,

And,
• angle POQ = angle SOR { vertically opposite angles }

Similarly,
• angle POS = angle QOR = 90°




Hence,
• PQ = PS = SR = RQ

• angle POQ = angle SOR = angle QOR = angle POS = 90°


Hence, the formula quadrilateral is a Rhombus.
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