Question 161720
Given:
{{{P}}}, {{{Q}}}, {{{R}}} and {{{S }}}are the{{{mid-points }}}of {{{AB}}}, {{{BC}}}, {{{CD}}} and {{{DA}}} respectively.
To prove: 
{{{PQRS}}} is {{{a}}} {{{rhombus}}}. 

Proof:

Consider triangle {{{BAC }}}
{{{PQ}}}|| {{{AC}}} and {{{PQ = (1/2)AC}}}….(1)

(In a triangle the segment joining the mid-points of 
two sides are parallel and equal to third side)
Consider triangle {{{ADC }}},

{{{SR}}}|| {{{AC}}} and {{{SR = (1/2)AC}}}….(2)



From (1) and (2),
{{{PQ}}}|| {{{SR}}} and {{{PQ=SR}}}

{{{PQRS}}} is a parallelogram ……………….(3)

{{{AD = BC}}}….. (opposite sides of a rectange)

So
{{{(1/2)AD =(1/2)BC}}}…..

i.e. {{{AS = BQ }}}

Consider triangle {{{ APS }}}, and triangle {{{ BPQ }}},

{{{AP = BP}}}…. ({{{P}}} is the mid-point of {{{AB}}})
{{{AS = BQ}}}

angle {{{SAP}}} =  {{{PBQ}}} = {{{90degrees}}}

So, {{{triangle}}} {{{ APS }}} is {{{congruent}}} to triangle {{{ BPQ }}}.... ({{{SAS}}} congruency condition)
 
{{{ PS = PQ}}} ………………………..(4) 
From (3) and (4),
{{{PQRS}}} is a parallelogram in which {{{PS = PQ }}}.
 {{{PQRS}}} is a {{{rhombus}}} .