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)`

Explanation

We know that distances of the vertices of a triangle from the points of contract to the incircle are `s - a, s- b, s- c`. Thus,
`alpha = s -a, beta = s -b, and gamma = s -c`,
Now, `alpha beta gamma = (s-a) (s-b) (s-c)`
and `alpha + beta + gamma = 3s - a - b - c = s`
`rArr (alpha beta gamma)/(alpha + beta + gamma) = ((s-a) (s-b) (s-c))/(s)`
`= (s(s-a) (s-b) (s-c))/(s^(2))`
`= (Delta^(2))/(s^(2))`
`= r^(2)`