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

Given - In $\Delta ABC,\angle {90}^0$ which is inscribed in a circle and O is the incentre of the incircle of $\Delta$ ABC D and d are the diameters of circumcircle and incircle of $\Delta$ ABC.

Construction- Joint OL, OM and ON.

In circumcircle of $\Delta$ ABC,

$\angle B={90}^0$ (given)

$\therefore$ AB is the diameter of circumcircle i.e., AB = D.

Let radius of incircle = r

$\therefore$ OL =OM=ON =r

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

$\therefore$ BL=OM =r

Similarly we can prove that :

AM =AN and CL =CN = R (radius)

(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= 2 AC+2r= 2D+d

Q.E.D.