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.

Answer 1

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