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  • In Fig., CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY)

In Fig., CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY)

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Solution.

Given: CD\parallel AE and CY \parallel BA
To prove: ar (\triangle CBX) = ar(\triangle AXY)
Proof: We know that, triangles on the same base and between the same parallels have equal area.
Consider CY\parallel BA
\triangle ABYand \triangle ABC both lie on the same base AB and between the same parallels CY and BA.
ar(\triangle ABY) = ar(\triangle ABC)                    …(1)
Now, ar(\triangle ABY) = ar(ABX) + ar(AXY)
and, ar(\triangle ABC) = ar(ABX) + ar(CBX)
Putting the above values in eq. (1)
\Rightarrow ar(ABX) + ar(AXY) = ar(ABX) + ar(CBX)
\Rightarrow ar(AXY) = ar(CBX)
Hence proved

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