Q: 13     
ABCD is a trapezium with \small AB\parallel DC. A line parallel to AC intersects AB at X and BC at Y. Prove that \small ar(ADX)=ar(ACY).

 [Hint: Join CX.]

Answers (1)


We have a trapezium ABCD, AB || CD
XY ||AC meets AB at X and BC at Y. Join XC

Since \DeltaADX and \DeltaACX lie on the same base CD and between same parallels AX and CD
Therefore, ar(\DeltaADX) = ar(\DeltaACX)..........(i)
Similarly ar(\DeltaACX) = ar(\DeltaACY).............(ii) [common base AC and AC || XY]
From eq (i) and eq (ii), we get

ar(\DeltaADX) = ar (\DeltaACY)

Hence proved.

 

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