prove that area of the quadrilateral formed by joining the mid points of the sides of a parallelogram is half the area of the parallelogram
Answer :
Let P , Q , R , S are the mid point of parallelogram ABCD .
And
Diagonals of parallelogram ABCD meet at " O " , And we join OP , OQ , OR and OS .
And
We know AB = CD , AB | | CD and BC = DA , BC | | DA ( Sides of parallelogram )
And
AP = PB = AB
And
RC = RD = CD And we know AB = CD , SO
AP = PB = RC = RD
Similarly
BQ = QC = DS = SA
We also know diagonal of parallelogram bisect each other , SO " O " is mid point of AC and BD .
Now In ABC
O is mid point of AC and Q is mid point of BC , So
Line OQ is parallel to AB , As we know from basic proportionality theorem , SO
OQ | | AB | | CD Then
OQ | | RC ( RC is a part of line CD ) ------------- ( 1 )
So ,
So,
OQ = RC ----------------( 2 )
Similarly we can show
QC | | OR , QC = OR ---------------- ( 3 )
So, From equation 1 , 2 and 3 , we get
OQCR is a parallelogram , So OF = FC As we know diagonals of parallelogram bisect each other .
So ,
Area of OFQ = Area of CFQ --------------- ( 4 ) ( As they have same base ( OF = FC ) And same height as they have same vertex " Q " )
Similarly
Area of OEQ = Area of BEQ --------------- ( 5 )
Add equation 4 and 5 , we get
Area of OFQ + Area of OEQ = Area of CFQ + Area of BEQ
Area of OEQF = Area of CFQ + Area of BEQ --------------- ( 6 )
Similarly
Area of OFRG = Area of CFR + Area of DGR --------------- ( 7 )
And
Area of OGSH = Area of DGS + Area of AHS --------------- ( 8 )
And
Area of OHPE = Area of AHP + Area of BPE --------------- ( 9 )
Now we add equation 6 , 7 , 8 and 9 and get
Area of OEQF + Area of OFRG + Area of OGSH + Area of OHPE = Area of CFQ + Area of BEQ + Area of CFR + Area of DGR + Area of DGS + Area of AHS + Area of AHP + Area of BPE
Area of quadrilateral PQRS = Area of QCR + Area of RDS + Area of SAP + Area of PBQ
Area of quadrilateral PQRS = Area of parallelogram ABCD - Area of quadrilateral PQRS
2 ( Area of quadrilateral PQRS ) = Area of parallelogram ABCD
Area of quadrilateral PQRS = ( Hence proved )