Q19 of 25 Page 9

In a quadrilateral ABCD diagonals AC and BD intersect at O such that OB=OD. If AB=CD, show that ar(Δ AOD)=ar(Δ BOC)

Construction:

Draw DP⊥ AC and BQ⊥ AC.



Proof:


In ΔDOP and Δ BOQ


∠DOP = ∠BOQ (vertically opposite)


OD = OB


∠DPO = ∠BQO (each 90°)


By AAS congruency,


ΔDOP ≅ Δ BOQ


⇒ DP = BQ


And ar (DOP)=ar(Δ BQO) …..(1)


Now consider ΔAPD and ΔCQB,


AD = BC


DP = BQ


∠DPO = ∠BQO (each 90°)


By RHS congruency,


ΔAPD ≅ ΔCQB


∠APD = ∠BQC


⇒ ar ΔAPD = ar ΔCQB ….. (2)


From (1) and (2),


ar (DOP) + ar ΔAPD =ar(Δ BQO) + ar ΔCQB


⇒ ar(Δ AOD) = ar(Δ BOC)


Hence Proved


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