If a,b and c are the …
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Sia ? 6 years, 6 months ago
Let a,b and c be the sides of right angled {tex} \triangle ABC{/tex}, such that BC = a, CA = b and AB = c.
Let the circle touches the sides BC, CA, AB at D, E and F, respectively.
Then, {tex}AE=AF{/tex}
and {tex}BD=BF{/tex}
and {tex}CD=CE{/tex} [tangents drawn from an external point are equal in length]
In quadrilateral OECD each angle is right angle .So, OECD is a square therefore, we have
{tex}OE=EC=CD=OD=r{/tex}
{tex}\therefore AF=AE=b-r\;{/tex} and {tex}BF=BD=a-r{/tex}
{tex}\Rightarrow AF+BF=(b-r)+(a-r){/tex}
{tex}\Rightarrow AB=b-r+a-r{/tex}
{tex}\Rightarrow c=a+b-2r{/tex}
{tex}\Rightarrow\ 2r=a+b-c{/tex}
{tex}\therefore r=\frac{a+b-c}{2}{/tex}
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