By applying Pythagoras Theorem in ∆ADB, we get,

AB^{2} = AD^{2} + DB^{2}…………… (i)

Again, by applying Pythagoras Theorem in ∆ACD, we get,

AC^{2} = AD^{2} + DC^{2}

AC^{2} = AD^{2} + (DB + BC) ^{2}

AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2DB × BC

From equation (i), we can write,

AC^{2} = AB^{2} + BC^{2} + 2DB × BC

Hence, proved.

Answered by Abhisek | 1 year agoIn an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm^{2}. Find ar(∆COD)

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