If the distances of the vertices of a triangle ABC from the points of contacts of the incircle with sides are $α, β$ and $γ$ then prove that $r^{2} = \frac{α β γ}{α + β + γ}$

Open in App

Solution

We know that distances of the vertices of a triangles from the points of
contact of the incircle is $s - a, s - b, s - c$

Thus, $α = s - a β = s - b$ and $γ = s - c$

Now
$α β γ = (s - a) (s - b) (s - c)$
and $α + β + γ = 3 s - a - b - c = s$
$∴ \frac{α β γ}{α + β + γ} = \frac{(s - a) (s - b) (s - c)}{s}$
$= \frac{s (s - a) (s - b) (s - c)}{s^{2}}$ $= \frac{Δ^{2}}{s^{2}}$ $= r^{2}$

Suggest Corrections

Similar questions

α, β, γ

α, β, γ

x^{3} - 2 x^{2} - x + 2 = 0

(α, β, γ), (β, γ, α), (γ, α, β)

α

β

γ

α \times s i n β - c o s γ = c o s α c o s β

α, β, γ

x^{3} - 3 p x^{2} + 3 q x - 1 = 0

(α, \frac{1}{α})

(β, \frac{1}{β})

(γ, \frac{1}{γ})

{2 x}^{3} + {3 x}^{2} + 5 x + 6 = 0

α, β, γ

α + β + γ, α, β + β γ + γ α

α β γ

Join BYJU'S Learning Program