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In a triangle C=90. If r is the in-radius and R is the circum-radius,then prove that (2r+R) is equal to a a+b-c/2.
Let the perps from incentre I to BC, AB and AC be D, E and F resp. The FCDI is a square of side r. Then AF = AE = b-r (tangents from A to the incircle) and similarly, BD = BE = a-r. Then c= AB = AD+BD = a+b-2r. Now AB is the diameter of the circumcircle, and hence we have c = 2R, so that R = c/2 Thus 2R = a+b – 2r or 2r+R = a+b – R = a+b – c/2
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