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Question
Prove that 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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Solution
Join OL, OM and ON.
Let D and d be the diameter of the circumcircle and incircle.
And let R and r be the radius of the circumcircle and incircle.
In circumcircle of ΔABC,
∠B = 90°
Therefore, AC is the diameter of the circumcircle i.e. AC = D
Let radius of the incircle = r
∴ OL = OM = ON = r
Now, from B, BL, BM are the tangents to the incircle.
∴ BL = BM = r
Similarly,
AM = AN and CL = CN = R
(Tangents from the point outside the circle)
Now,
AB + BC + CA = AM + BM + BL + CL + CA
= AN + r + r + CN + CA
= AN + CN + 2r + CA
= AC + AC + 2r
= 2AC + 2r
= 2D + d
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