# 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.]

M manish

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

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

ar($\Delta$ADX) = ar ($\Delta$ACY)

Hence proved.

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