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Two chords AB and CD of a circle interse...

Two chords AB and CD of a circle intersect at a point outside the circle. Prove that
`(a) Delta PAC~ Delta PDB" " (b) PA*PB= PC*PD`.

Text Solution

Verified by Experts

Clearly ABCD is a cyclic quadrilateral.
`:. angle1+ angle2 =180^(@)" "...(i)`
In `Delta PAC and Delta DPB`, we have
`angle PAC=angle DPB`[ common]
`angle PAC = angle PDB` [ each equal to `(180^(@)-angle)`]
[ Note `angle PAC+ angle 1=180^(@)` (linear pair)
and `angle PDB+ angle1=180^(@)` [ using]
`:. Delta PAC~Delta PDB` [ by AA-similarity]
And so, `(PA)/(PD)=(PC)/(PB) rAr PA*PC*PD`.
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