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Question

# In ΔABC, if the incircle touch the sides BC,CA,AB at D,E,F, respectively, then BD+CE+AF equals

A
s
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B
2s
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C
s2
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D
None of these
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Solution

## The correct option is A sWe know that the length of tangent drawn from an external point to a circle are equal.∴AF=AE------------(i)BD=BF------------(ii)CE=CD------------(iii)Add (i), (ii), and (iii), we getAF+BD+CE=AE+BF+CD------------(iv)Perimeter of △ABC=AB+BC+CA=(AF+BF)+(BD+CD)+(CE+AE)=(AF+BD)+(BD+CE)+(CE+AF)-------from (i), (ii), and (iii)=2(AF+BD+CE)AF+BD+CE=12(Perimeter of △ABC)We know that s is the semi-perimeter.AF+BD+CE=s

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