Prove that the distance between the circumcenter and the incenter of triangle ABC is`sqrt(R^2-2R r)`

Distance Between Circumcenter and incenter

In A B C ,l e tR=c i r c u m r a d i u s ,r=in r a d i u sdot If r is the distance between the circumcenter and the incenter, the ratio R/r is equal to sqrt(2)-1 (b) sqrt(3)-1 sqrt(2)+1 (d) sqrt(3)+1

The distance between the circumcentre and orthocenter of a Delta ABC is R sqrt(1-lambda cos A*cos B*cos C), then lambda=

Prove using vectors that the distance of the circumcenter of the Delta ABC form the centroid is sqrt(R^(2) - (1)/(9)(a^(2) + b^(2) + c^(2))) where R is the circumradius.

If R be the circum radius and r the in radius of a triangle ABC, show that Rge2r

Let ABC be an equilateral triangle with side length a. Let R and r denote the radii of the circumcircle and the incircle of triangle ABC respectively. Then, as a function of a, the ratio R/r

Prove that the distance of the incentre of Delta ABC from A is 4R sin B/2 sin C/2.

In triangle ABC ,the line joining the circumcenter and incenter is parallel to side AC, then cos A+cos C is equal to -1 (b) 1 (c) -2(d)2