I don't really understand it, we shall prove that: $\frac{a}{{(p-b)}\cdot{(p-c)}}=\frac{2}{r\cdot{tg{\alpha}}}$ $ \leftrightarrow$ $\frac{a{\cdot{p}\cdot{(p-a)}}}{S^2}=\frac{2}{r\cdot{tg{\alpha}}}$, or $\frac{p-a}{h_a}=\frac{1}{tg{\alpha}}$ which is not quite true. Sorry if I missed something

Sailor wrote: I don't really understand it, we shall prove that: $ \frac {a}{{(p - b)}\cdot{(p - c)}} = \frac {2}{r\cdot{tg{\alpha}}}$ $ \leftrightarrow$ $ \frac {a{\cdot{p}\cdot{(p - a)}}}{S^2} = \frac {2}{r\cdot{tg{\alpha}}}$, or $ \frac {p - a}{h_a} = \frac {1}{tg{\alpha}}$ which is not quite true. Sorry if I missed something I think $ p$ is the perimeter, so $ p = a+b+c$.

if b = c, ∠ APB=α = pi / 2. tga = INF

Obviously, there's something wrong in the statement. since the left side of the equation is always positive, while the right side will be negative when $\alpha$ is obtuse.

Omid Hatami wrote: Suppose in triangle $ABC$ incircle touches the side $BC$ at $P$ and $\angle APB=\alpha$. Prove that : \[\frac1{p-b}+\frac1{p-c}=\frac2{rtg\alpha}\] I guess it must be $\frac1{p-b}-\frac1{p-c}=\frac2{rtg\alpha}$

Good job K.N! I think that is what it should really be.

Suppose in triangle $ABC$ incircle touches the side $BC$ at $P$ and $\angle APB=\alpha$. Prove that : \[\frac1{p-b}-\frac1{p-c}=\frac2{rtg\alpha}\]correct wording