In the adjoining figure, PQRS is a parallelogram. A and B are the mid-points of PQ of SR respectively. If PS= BR, then prove that quadrilateral ADBC is a reactangle.

Explanation

Since `squarePQRS` is a parallelogram
`therefore" "PQ=SR and PS=RQ`
`implies" "1/2PQ=1/2SR`
`implies" "PA"||"SBand AQ=BR" "...(1)" "(because"A and B are mid-points")`
`AlsoPS=BR=SB=PA" "...(2)`
Now since `PQ"||"SR`
`implies" "PA"||"SB" "...(3)`

`therefore` From (1) and (2), we get
`squarePABS` is a parallelogram
But since PA = PS = SB
`therefore squarePABS` is a rhombus.
Similarly `squareAQRB` is a rhombus
`{:(therefore,angle1=90^(@)),(and,angle2=90^(@)):}}" "{:((therefore"Diagonals of a rhombus biscet"),("each other perpendicularly.")):}`
Now, we know that diagongal PB bisects `angleP and angleQ` bisects `angleQ.`
(we can prove easily by proving `DeltaPAC and DeltaPSC` coagruent)
`But (angle3+angle4+(angle5+angle6=180^(@)" "("sum of oco-interior angles")`
`implies" "2angle4+2angle5=180^(@)`
`implies" "angle4+angle5=90^(@)`
`implies" "angle7=90^(@)`
Now, in `squareADBC,angle1=angle2=angle7=90^(@)" "("angle sum property")`
Each of the 4 angle of a quadrilateral ABCD is `90^(@).`
So, `squareADBC` is a reactangle.