**(i)** In ∆PAC and ∆PDB,

∠P = ∠P (Common Angles)

As we know, exterior angle of a cyclic quadrilateral is

∠PCA and ∠PBD is opposite interior angle, which are both equal.

∠PAC = ∠PDB

Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)

**(ii)** We have already proved above,

∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.

Therefore,

\( \dfrac{AP}{DP}=\dfrac{PC}{PB}=\dfrac{CA}{BD}\)

\( \dfrac{AP}{DP}=\dfrac{PC}{PB}\)

AP. PB = PC. DP

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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A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

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