Question

solve it fast pleas
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Answer 1

Answer:

(i)

Opposite angles of a cyclic quadrilateral are supplementary.

So ∠PBD = 180° - ∠ACD = ∠PCA.

Also, ∠BPD = ∠CPA is common to the two triangles PAC and PDB.

Therefore ΔPAC is similar to ΔPDB   ( by AA rule ).

(ii)

Since ΔPAC is similar to ΔPDB,

PA / PD = PC / PB

=> PA . PB = PC . PD