4 years ago

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

1 answer:

8 0

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

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$\bf \begin{cases}
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\\\\\\
\boxed{f(18)=20}\\\\
-------------------------------\\\\
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\\\\\\
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4 years ago