SHOW THAT QUADRILATERAL FORMED BY JOINING THE MID POINTS OF ADJACENT SIDES OF A RECTANGLE IS A RHOMBUS

Asked by  | 12th Mar, 2013, 12:54: AM

Expert Answer:

Given :- ABCD is a rectangle and P,Q,R,S are their midpoints.

To Prove:- PQRS is a rhombus.

Proof:- In Tri. ABC, P and Q are the mid points . 

So, PQ is parallel AC and PQ = 1/2 *AC (the line segment joining the mid points of 2 sides of the triangle is parallel to the third side and half of the third side) 

Similarly RS is parallel AC and RS = 1/2* AC

Hence, both PQ and RS are parallel to AC and equal to 1/2*AC. Hence, PQRS is a parallelogram

In triangles APS & BPQ,

AP=BP (P is the mid point of side AB)

angle PAS = angle PBQ (90 degree each)

and, AS = BQ (S and Q are the mid points of AD and BC respectively and since oppsite sides of a rectangle are equal, so their halves will also be equal)

triangle APS is congruent to triangle BPQ (By SAS)

So, PS=PQ (By CPCT)

PQRS is a parallelogram in which adjacent sides are equal, ...PQRS is a rhombus

Answered by  | 12th Mar, 2013, 07:45: AM

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