Question

The semi perimeter of ∆ABC = 28cm then AF+BD+CE=​

Answer 1

Answer:

Since length of tangents from an external point to a circle are equal

Therefore AF=AD

DB=BE

CF=CE

Since semiperimeter of ΔABC=28cm

⇒ perimeter of ΔABC=56cm

⇒AB+BC+AC=56

(AF+DB)+(DB+CE)+(AF+CE)=56

2(AF+BD+CE)=56

AF+BD+CE=28cm

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