The sides a, b, c of a right triangle, where c is the hypotenuse, are circumscribing a
circle. Prove that the radius r of the circle is given by r =(a+b+c)/2
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Answered by
26
Answer:
Let the circle touches the sides BC,CA,AB of the right triangle ABC at
D,E and F respectively. Where,
BC=a
CA=b
AB=c.
Then, AE=AF and BD=BF. Also CE=CD=r.
b-r =AF, a-r=BF
r = (a+b+c)/2
Hence proved.
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Answered by
10
Answer:
r=a+b-c/2
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