ABCD is a trapezium with Ab parallel DC a line parallel to AC intersect ab at X and PB at y. prove that ar (ADx)=ar (Acy )
lumbi:
where is point p
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It can observe that ADX & ACX lie on same base AX and between same parallel AB and DC
ar.(ADX)=ar.(ACX)
ACY and ACX lie on same base AX and between same parallel AC and XY.(1)
ar.(ACY)=ar.(ACX). (2)
From 1 and 2,
ar.(ADX)=ar.(ACY)
ar.(ADX)=ar.(ACX)
ACY and ACX lie on same base AX and between same parallel AC and XY.(1)
ar.(ACY)=ar.(ACX). (2)
From 1 and 2,
ar.(ADX)=ar.(ACY)
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Yaa this fig is correct
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