Q. 84.4( 71 Votes )

In Fig. 6.62, two

Answer :

(i) In triangle PAC and PDB

PAC + CAB = 180o (Linear pair)


CAB + BDC = 180O (Opposite angles of a cyclic quadrilateral are supplementary)


Hence,


PAC = PDB


Similarly,


PCA = PBD


Hence,


Δ PAC ~ Δ PDB


(ii) Since the two triangles are similar, so



Or,


PA * PB = PC * PD


Hence, proved


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