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In the given figure AD||BC, Prove ar(triangleGCD)=ar(triangleABG)

In the given figure AD||BC, Prove ar(triangleGCD)=ar(triangleABG)-example-1

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triangla ADC and ADB are on same base n between same parallel lines so they are equal in area n they have common triangle ADG if we subtract this triangle from both triangle ADC n ADB we can find triangle GCD n triangle ABG. equal

In the given figure AD||BC, Prove ar(triangleGCD)=ar(triangleABG)-example-1
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User Kdeez
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