If the distances of the vertices of a triangle from the point of contact of the incricle with the sides be `alpha, beta, gamma` then r is equal to (where r=inradius)

If the distances of the vertices of a triangle =ABC from the points of contacts of the incercle with sides are alpha,beta and gamma then prove that r^(2)=(alpha beta gamma)/(alpha=beta+gamma)

If the distances of the vertices of a triangle =ABC from the points of contacts of the incercle with sides are alpha,betaa n dgamma then prove that r^2=(alphabetagamma)/(alpha+beta+gamma)

If the distances of the vertices of a triangle =ABC from the points of contacts of the incercle with sides are alpha,betaa n dgamma then prove that r^2=(alphabetagamma)/(alpha+beta+gamma)

If the distances of the vertices of a triangle =ABC from the points of contacts of the incercle with sides are alpha,betaa n dgamma then prove that r^2=(alphabetagamma)/(alpha+beta+gamma)

In a Delta ABC, D,E and F respectively are the points of contact of the incircle with the sides AB, BC and CA such that AD =alpha, BE= beta and CF =gamma," then "(alpha beta gamma)/(alpha +beta +gamma)=

If alpha , beta, gamma are the lengths of the tangents from the vertices of a triangle to its incircle. Then

In a triangle ABC, let /_C=pi//2 . If r is the inradius and R is the circum-radius of the triangle, then 2(r+R) is equal to :

If the centroid of a triangle with vertices (alpha, 1, 3), (-2, beta, -5) and (4, 7, gamma) is the origin then alpha beta gamma is equal to …..........

If the centroid of a triangle with vertices (alpha, 1, 3), (-2, beta, -5) and (4, 7, gamma) is the origin then alpha beta gamma is equal to …..........