Theorem :If from one external point, two tangents are drawn to a circle then they have equal tangent segments.
Two tangents DB and BE are drawn to a circle from one point i.e.B
So, Using theorem
DB=BE =10 cm
So, Using theorem
CF = CE = 15 cm
So, using theorem
AD =FA = 8 cm
Now AC = FA+CF = 8+15=23 cm
CB =BE+EC = 10+15=25 cm
AB = AD+DB=8+10=18 cm
Perimeter of triangle = Sum of all sides
=AB +CB+AC
= 18 + 25 +23
= 56 cm
Hence the perimeter of ΔACB is 56 cm
