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  • Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, prove that BD = s – b.

Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, prove that BD = s – b.

Answers (1)

Solution
Given : BC = a, CA = b, AB = c

Here, AF = AE = Z1 (\mathbb{Q}  tangents drawn from an external point to the circle are equal in length)

CE=CD=Z_{2} 
BD=BF=Z_{3} 
Here AB + BC + CA = c + a + b       
(AB + FB) + (BC + DC) + (CE + EA) = a + b + c
\left ( AB+FB \right )+\left ( BC+DC \right )+\left ( CE+EA \right )= a+b+c
\left ( Z_{1} +Z_{3}\right )+\left ( Z_{3} +Z_{2}\right )+Z_{2} +Z_{1}= a+b+c
2\left ( Z_{1}+Z_{2} +Z_{3}\right )= a+b+c
2\left ( Z_{1}+Z_{2} +Z_{3}\right )= 2S
2\left ( Z_{1}+Z_{2} +Z_{3}\right )= 2S
\left ( a+b+c= 2s= perimeter\, of\, \bigtriangleup ABC \right )
Z_{1}+Z_{2}+Z_{3}= s
Z_{3}= s-\left ( Z_{1} +Z_{2}\right )
-\Rightarrow BD= s-b            \left [\because b= CE+EA= Z_{1}+Z_{2} \right ]
Hence Proved         

Posted by

infoexpert27

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