#### 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.

#### 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

Is there an error in this question or solution?

Solution 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. Concept: Arc and Chord Properties - Angle in a Semi-circle is a Right Angle.