Question

ABCD is a rectangle and P is the midpoint of AB.DP is produced to meet CB at Q. Prove that area of rectangle ABCD = area of triangle DQC

Answer 1

in triangle ADP and triangle PBQ

AP=BP. (Pis the mid point of AB)

<APD=<BPQ. (vertically opposite )

<DAP=<QBP=90°

triangle ADP and triangle PBQ are congruent (ASArule)

therefore ar(ADP) =ar(PBQ)

ar(ADP)+ar (DPBC)=ar(PBQ)+ ar (DPBC)

ar (ABCD). =ar (DQC