prove that the perimeter of a right angled triangle is equal to the sum of diameter of its incircle and twice diameter of it's circumcircle
ABC is a right angled triangle right angled at B.
Hence AC is diameter of the circumcircle as angle in a semi circle is a right angle.
O is the centre of incircle and BP and BQ are tangents
Hence BP = BQ (Tangents drawn from an external point to a circle are equal in length)
Similarly we have CQ = CR and AP = AR
Consider the quadrilateral PBQO in which ∠B = 90°
∠Q = ∠P = 90° [Since radius perpendicular to tangent]
Therefore, PBQO is a square.
Hence BP = BQ = OP = OQ → (1)
Let BP = BQ = OP = OQ = r (radius of incircle) → (2)
Consider, AB + BC + CA = AP + BP + BQ + CQ + AC
= AR + OP + OP + CR + AC
= (AR + CR) + 2OP + AC
= AC + 2OP + AC
= 2AC + 2r = 2AC + d (Diameter d = 2r)
= 2(Diameter of circumcircle) + (Diameter of incircle)
Thus the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
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