How prove $\angle ABC=60^o$ for triangle ABC if $BD = BF = AB-AC$ and $\frac{1}{{AD}}+\frac{1}{{CF}}=\frac{1}{{BD}}$? In acute triangle $ABC$ with $AB > AC$,we consider points $D$, $F$ (internal) of $AB$, $BC$ respectively, such that $BD = BF = AB-AC$.
If: $\frac{1}{{AD}}+\frac{1}{{CF}}=\frac{1}{{BD}}$, to how prove that $\angle ABC ={60^\circ}$?
 A: From $BD = BF = AB-AC$ you can deduce that $AD = AC$.
Let $r = BD = BF$ and $R = AC = AD$.
From the other equation we deduce that $CF = \displaystyle\frac{Rr}{R-r}$.
Using the law of cosines:
$$(R+r)^2 + \left(r + \frac{Rr}{R-r}\right)^2 - R^2 = 2(R+r)\left(r + \frac{Rr}{R-r}\right)\cos B$$
Or equivalently:
$$(R+r)^2(R-r)^2 + (2Rr-r^2)^2 - R^2(R-r)^2 = 2(R+r)(R-r)(2Rr-r^2)\cos B $$
So:
$$\frac{(R+r)^2(R-r)^2 + (2Rr-r^2)^2 - R^2(R-r)^2}{2(R+r)(R-r)(2Rr-r^2)} = \cos B \\ \Rightarrow \frac{(2Rr+r^2)(R-r)^2 + r^2(2R-r)^2 }{2r(R+r)(R-r)(2R-r)} = \cos B\\ \Rightarrow \frac{(2R+r)(R-r)^2 + r(2R-r)^2 }{2(R+r)(R-r)(2R-r)} = \cos B \\ \Rightarrow \frac{1}{2}·\frac{2r^3 - 4Rr^2 + R^2r +2R^2}{r^3-Rr^2-R^2r+2R^2} = \cos B$$
We want $\cos B = 1/2$. So we want the biq quotient to be $1$. That only can be if: $$r^3 - 3r^2R+2R^2r = 0 \Rightarrow r(r-2R)(r-R)=0$$
Now $r \neq 0$ because $r = AB - AC$ and $AB > AC$.
So we have to prove that either $R = r$ or $2R = r$
Note that $r = R$ is not possible because of the second equation:
$$\frac{1}{R} + \frac{1}{CF} = \frac{1}{r}$$
So if $r = R$ then $\displaystyle\frac{1}{CF} = 0$ wich cannot be.
The same reasoning goes for $2R = r$:
$$\frac{1}{R} + \frac{1}{CF} = \frac{1}{2R}$$
What means that $CF = -R$ Wich is impossible. Note that since $CF \displaystyle\frac{Rr}{R-r}$ then we would have $Rr = (r-R)R$ what would mean that $R = 0$ so we wouldn't have a triangle.
Therefore, $\angle ABC \neq 60^\circ$
