If $a, b, c$ are the sides of a triangle and $ka, kb, kc$ are the sides of a similar triangle inscribed in the former and $F$ be the angle between the sides $a$ and $kc$, prove that $2k\cos(F)=1$.
I tried to apply the sine rule to triangle $ADF$, obtainin $AD = \dfrac {kc \sin(A+F)} {\sin A}$. The sine rule for triangle $BDE$ gives $BD = \dfrac {kb \sin(A+F-B)} {\sin B}$. But $BD+DA =c$.
I need help from here. Please not that this is not a homework question.
We know that $\triangle ABC\sim \triangle EFD$
So $\frac{AB}{EF}=\frac{BC}{FD}=\frac{CA}{DE}=\frac1k$
$\frac{a}{EF}=\frac{c}{FD}=\frac{b}{DE}=\frac1k$
Using the law of cosines in $\triangle ADF$, we get $$AF^2=AD^2+DF^2-2AD.DF.\cos\theta$$ $$2AD.kc\cos\theta=AD^2+k^2c^2-AF^2$$ $$2k\cos\theta=\frac{AD^2}{AD.c}+\frac{k^2c^2}{AD.c}-\frac{AF^2}{AD.c}$$ $$2k\cos\theta=\frac{AD}{c}+\frac{k^2.c}{AD}-\frac{AF^2}{AD.c}$$ $$...$$ Now show that the RHS is equal to $1$.