A geometry problem - easy with trigonometry, harder without it

Let $D$ be the point where the bisector of angle $C$ hits side $AB$.

Then we get

• $\angle BCD=55^\circ$$\\[4pt]$
• $\angle BDC=70^\circ$$\\[4pt]$
• $\angle ACD=55^\circ$$\\[4pt]$
• $\angle ADC=110^\circ$

hence $\Delta BDC$ is isosceles, with $|DB|=|DC|$, and $\Delta ABC$ is similar to $\Delta ACD$.

The factor for scaling side lengths of $\Delta ABC$ to corresponding side lengths of $\Delta ACD$ is $$\frac{|AC|}{|AB|}=\frac{b}{c}$$ hence, we get \begin{align*} |DC|&=\left(\frac{b}{c}\right)a\\[4pt] |AD|&=\left(\frac{b}{c}\right)b\\[4pt] \end{align*} But then also \begin{align*} |AD|&=c-|DB|\\[4pt] &=c-|DC|\\[4pt] &=c-\left(\frac{b}{c}\right)a\\[4pt] \end{align*} so \begin{align*} &c-\left(\frac{b}{c}\right)a=\left(\frac{b}{c}\right)b\\[4pt] \implies\;&c^2-ab=b^2\\[4pt] \implies\;&c^2=ab+b^2 \end{align*}

All that matters is that $\angle C = 2 \angle B$:

$$\triangle ABC \sim \triangle AB^\prime B \qquad\to\qquad \frac{c}{b}=\frac{a+b}{c}$$