Question

If $\alpha ,\beta ,\gamma$ are the distances of the vertices of a triangle ABC from the corresponding points of contacts with the incircle, then

• A. $s=\alpha +\beta +\gamma$
• B. $r^2=\frac{\alpha \beta \gamma }{\alpha +\beta +\gamma }$
• C. ${\Delta }^2=(\alpha +\beta +\gamma )\alpha \beta \gamma$
• D. $2s=\alpha +\beta +\gamma$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $s=\alpha +\beta +\gamma$
B $r^2=\frac{\alpha \beta \gamma }{\alpha +\beta +\gamma }$
C ${\Delta }^2=(\alpha +\beta +\gamma )\alpha \beta \gamma$

The perimeter of the triangle is

$2(\alpha +\beta +\gamma )$

$\therefore s=\alpha +\beta +\gamma$

We know that,

$r=\frac{\Delta }{s}$

$\therefore r^2=\frac{{\Delta }^2}{s^2}$ $\dots$

${\Delta }^2=s(s-a)(s-b)(s-c)$

$\therefore r^2=\frac{(s-a)(s-b)(s-c)}{s}$

$s-a=\alpha +\beta +\gamma -(\alpha +\beta )=\gamma$

$s-b=\beta$

$s-c=\gamma$

substitute them in $\left(1\right)$

${\Delta }^2=(\alpha +\beta +\gamma )\alpha \beta \gamma$

$\therefore r^2=\frac{\alpha \beta \gamma (\alpha +\beta +\gamma )}{(\alpha +\beta +\gamma {)}^2}=\frac{\alpha \beta \gamma }{\alpha +\beta +\gamma }$