Given, ABC and AMP are two right triangles, right angled at B and M respectively.

**(i)** In ΔABC and ΔAMP, we have,

∠CAB = ∠MAP (common angles)

∠ABC = ∠AMP = 90° (each 90°)

ΔABC ~ ΔAMP (AA similarity criterion)

**(ii)** As, ΔABC ~ ΔAMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal,

Hence, \( \dfrac{CA}{PA}=\dfrac{BC}{MP}\)

Answered by Sudhanshu | 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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