The perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
True
Given - In ΔABC,∠900 which is inscribed in a circle and O is the incentre of the incircle of Δ ABC D and d are the diameters of circumcircle and incircle of Δ ABC.
Construction- Joint OL, OM and ON.
In circumcircle of Δ ABC,
∠B=900 (given)
∴ AB is the diameter of circumcircle i.e., AB = D.
Let radius of incircle = r
∴ OL =OM=ON =r
Now from B, BL, BM are the tangents to the incircle
∴ 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.