Question

ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC

Answer 1

Answer:

In △ABP and △QCP,

∠BAP=∠PQX [ Alternate angles as AB∥DC ]

∠BPA=∠QPC [ Vertically opposite angles ]

∴ △ABP∼△QCP [ By AA similarity ]

⇒  

QC

AB

​  

=  

CP

BP

​  

=  

QP

AP

​  

 [ Corresponding sides of similar triangles ]

⇒  

QC

AB

​  

=  

CP

BP

​  

 

⇒ QC×BP=AB×BP

⇒ AB×BP+CQ×BP=AB×CP+AB×BP [Adding AB×BP on both sides ]

⇒ BP(AB+CQ)=AB(CP+BP)

⇒ BP(DC+CQ)=AB×CB

⇒ BP×DQ=AB×BC

∴ Rectangle obtained by BP and DQ = Rectangle obtained by AB and BC

Step-by-step explanation:

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