a,b,c are the sides of right triangle where c is hypotenuse a circle with radius r touches the sides of triangle prove that r=a+b+c/2
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Since tangents drawn from external points are equal. So AD=AF
BD=BE
CE=CF
Similarly Let EB=BD=radius or we can say r
So, c=AF+FC
=AD+CE
=(AB-BD) + (CB-EB)
= a-r +b-r
So, c= a-r +b-r
2r=a+b-c
So, r=a+b-c/2
Hence proved. Hope My answer will help you. Please select My answer as brainliest if it Actually does helped u.
BD=BE
CE=CF
Similarly Let EB=BD=radius or we can say r
So, c=AF+FC
=AD+CE
=(AB-BD) + (CB-EB)
= a-r +b-r
So, c= a-r +b-r
2r=a+b-c
So, r=a+b-c/2
Hence proved. Hope My answer will help you. Please select My answer as brainliest if it Actually does helped u.
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