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.

Let D and d be the diameter of the circumcircle and incircle.

Also, let R and r be the radius of the circumcircle and incircle.

Now, in circumcircle of ∆ABC,

AC is the diameter of the circumcircle i.e. AC = D

From figure,

OL = OM = ON = r

As, tangents from an exterior point to a circle are equal.

Now, from B, BL and BM are the tangents to the incircle.

So, BL = BM = r

Similarly,

AM = AN ………..(1)

CL = CN ………..(2)

Perimeter of ∆ABC = AB + BC + CA.

⇒ AB + BC + CA = AM + BM + BL + CL + CA

= AN + r + r + CN + CA …………[From (1) and (2)]

= AN + CN + 2r + CA

= AC + AC + 2r

= 2AC + 2r

= 2D + d

Hence, proved 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.