Given : A parallelogram ABCD in which bisectors of angles A,B, C,D intersect at P,Q,S to from a quadrilaateral PQRS.
To prove : PQRS is a rectangle.
Proof: Since, ABCD is a parallelogram.
Therefore, `AB"||"DC`
Now, `AB"||"DC` and transversal AD interects them at D and A respectively.
Therefore,
`" "angleA+angleD=180^(@)" "(because"sum of consective interior angles is"180^(@))`
`implies" "1/2angleA+1/2angleD=90^(@)`
`implies" "angleDAS+angleADS=90^(@)" "...(1)`
`" "(because"DS and AS are bisectors of" angleAand angleD "respectively")`
But , in `DeltaDAS` we have
`angleDAS+angleASD+angleADS=180^(@)" "(because"sum of angle of a triangle is"180^(@))`
`implies" "angle90^(@)+angleASD=180^(@)" "["using"(1)]`
`implies" "angleASD=90^(@)`
`implies" "anglePSR=90^(@)" "(becauseangleSADandanglePSR"are vertically opposite angle"thereforeanglePSR=angleASD)`
Similarly, we can prove that
`angleSEQ=90^(@),angleRQP=90^(@)andangleSPQ=90^(@)`
Hence, PQRS is a rectangle.
