Inscribed circle of an isosceles triangle

Question

In a triangle $ABC$, $AC = BC = 24$ and a circle with center $J$ is inscribed. If $CH$ is altitude $(CH\perp AB,H\in AB)$ and $CJ:CH=12:17$, then find the length of $AB$. Triangle $ABC$ is isosceles so $H$ is also the midpoint of $AB.$ So we can just find $AH$ (or $BH$) and double it to find $AB$. Let $CJ=12x, CH=17x\Rightarrow JH=5x$. I am not sure what to do from here. Thank you in advance!

Answer 1

Let $JH = r$, and $CH = h$. Then there are two formulas for the triangle's area: $$|\triangle ABC| = \frac{1}{2}ch = rs,$$ where $AB = c$ and $s = (a+b+c)/2$ is the semiperimeter. Since we know $a = b = 24$, it follows that $$ch = r(48+c).$$ But we are given that $CJ:CH = 12:17$, or $$\frac{h-r}{h} = 1 - \frac{r}{h} = \frac{12}{17}.$$ Thus $$c = \frac{r}{h}\left(48 + c\right) = \frac{5}{17}\left(48 + c\right),$$ which allows us to solve for $c$.

Answer 2

Since $J$ is incenter, $AJ$ is angle bisector in $\triangle HAC$. $AJ$ divides $CH$ in ratio of adjacent sides. $$CJ/JH = 12/5 = AC/AH \Rightarrow AH = 10 \Rightarrow AB=20$$