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# In Figure, two chords AB and CD intersect each other at the point P. Prove that : (i) $$\triangle$$ APC ~ $$\triangle$$ DPB (ii) AP . PB = CP . DP

Answer :

Firstly, let us join CB, in the given figure.

(i) In $$\triangle$$ APC and $$\triangle$$ DPB,
$$\angle$$ APC = $$\angle$$ DPB (Vertically opposite angles)
$$\angle$$ CAP = $$\angle$$ BDP (Angles in the same segment for chord CB)

Therefore,
$$\triangle$$ APC ~ $$\triangle$$ DPB (AA similarity criterion)

(ii) In the above, we have proved that $$\triangle$$ APC ~ $$\triangle$$ DPB

We know that the corresponding sides of similar triangles are proportional.
$$\frac{AP}{DP} = \frac{PC}{PB} = \frac{CA}{BD}$$
$$\frac{AP}{DP} = \frac{PC}{PB}$$
AP. PB = PC. DP
Hence, proved.