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    				| 1. | Let P, Q, R, S be respectively the midpoints of the sides AB, BC, CD and DA of quad. ABCD Show that PQRS is a parallelogram such that `ar("||gm PQRS")=(1)/(2)ar("quad. ABCD")`. | 
| Answer» Join AC and AR. In `triangleABC`, P and Q are midpoints of AB and BC respectively. `therefore` PQ||AC and PQ `=(1)/(2) AC`. In `triangleDAC`, S and R are midpoints of AD and DC respectively. `therefore` SR||AC SR`=(1)/(2) AC`. Thus, PQ||SR and PQ = SR. `therefore` PQRS is a ||gm. Now, median AR divides `triangle`ACD into two `triangle` of equal area. `therefore ar(triangleARD)=(1)/(2)ar(triangleACD)." "...(i)` Median RS divides `triangle`ARD into two `triangle` of equal area. `therefore ar(triangleDSR)=(1)/(2)ar(triangleARD)." "...(ii)` From (i) and (ii), we get `ar(triangleDSR)=(1)/(4)ar(triangleACD)`. Similarly, `ar(triangleBQP)=(1)/(4)ar(triangleABC)`. `therefore ar(triangleDSR)+ar(triangleBQP)=(1)/(4)[ar(triangleACD)+ar(triangleABC)]` `rArrar(triangle DSR)+ar(triangleBQP)=(1)/(4)ar["quad.ABCD"]." "...(iii)` Similarly, `ar(triangleCRQ)+ar(triangleASP)=(1)/(4)ar("quad. ABCD")." "...(iv)` Adding (iii) and (iv), we get `ar(triangleDSR)+ar(triangleBQP)+ar(triangleCRQ)+ar(triangleASP)` `=(1)/(2)ar("quad. ABCD")" "...(v)` But, `ar(triangleDSR)+ar(triangleBQP)+ar(triangleCRQ)+ar(triangleASP)` `+ar("||gm PQRS)=ar("quad. ABCD")." "...(vi)` Substracting (v) from (vi), we get `ar("||gm PQRS") =(1)/(2)ar("quad. ABCD")`. | |