Since, the length of tangents drawn from an external point to a circle are equal.
So,
AF = AE ..(i)
BD = BF ..(ii)
and CE = CD ..(iii)
Adding (i), (ii), (iii), we get
AF + BD + CE = AE + BF + CD ..(iv)
Now perimeter of
= AB + BC + AC
= AF + BF + BD + CD + AE + CE
= 2 (AF + BD CE) [using eq. iv]
AF + BD + CE =
But, AF + BD + CE + AE + DF + CD
Hence,
AF + BD + CE = AE + DF + CD
From a point P outside a circle, with centre O tangents PA and PB are drawn as shown in the figure, prove that
(i) ⇒AOP = ∠BOP
(ii) OP is the perpendicular bisector of AB.