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PP' and QQ' are 2 direct tangents to two circles intersection at A and B.The common chord is produced which intersect PP' at R and QQ' at S.Prove that i)RA=SB

ii)RS ^{2} =P'P ^{2}+ AB^{2}

1
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ii)RS ^{2} =P'P ^{2}+ AB^{2}

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So,

Likewise, The second circle with center O', RP' is the tangent like upper one and RAB is a secant to the circle.

So,

Equating both (1) and (2), we will get

Presently,

[As RS = RA + AB + BS]

[As we know from above RA = BS]

[by means of: RA + AB = RB]

[using the eq (1)]