parmenides51 wrote: Prove that if $\frac{AC}{BC}=\frac{AB + BC}{AC}$ in a triangle $ABC$ , then $\angle B = 2 \angle A$ . Hint: before you you even try to use Law of cosines or Law of sines to solves. draw $P$ on line $AC$ such that line $PB$ bisects $\angle ABC$. This makes $PB$ = $\frac{AC}{AB+BC}$($AB$) and $PC$ = $\frac{AC}{AB+BC}$($BC$), due to angle bisector theorem.

solve not solves

Since $\frac{AC}{BC}=\frac{AB+BC}{AC}$ let $A'$ be a point on $BC$ such that $A'B=AB$. Draw a circumcircle $k$ of $\triangle AA'B$. Therefore $AC$ is a tangent to $k$ at point $A$ and $\triangle AA'B$ is isosceles with $\angle AA'B=\angle BAA'$. $\angle B=2\angle AA'B$ as an exterior angle and $\angle A=\angle AA'B$ due to arc $AB$ and tangent property. Result follows.