In figure PA and PB are tangents from P to the circle with centre O. R is a point on the circle. Prove that : PC + CR = PD + DR
Since tangents from an external point of a circle are equal in length, therefore,
AD = AE, CD = CF and BE = BF.
Now, 2AD = AD + AD = AD + AE
= (AC + CD) + (AB + BE)
⇒ 2AD = AC + AB + (CD + BE)
⇒ 2AD = AC + AB + (CF + BF)
⇒ 2AD = AC + AB + BC