The base of an isosceles triangle; given the radius of the inscribed circle and perimeter

Question

On the sketch below $AC=CB$ and $OD=\dfrac{4}{10}CD$. If the perimeter of $\triangle ABC$ is $P_{\triangle ABC}=40$, find the length of the base $AB$.

Let $AC=BC=a$ and $AB=c$. Since the triangle is isosceles, $CD$ is also the altitude and the median. So we have $AD=BD=\dfrac12c$. How should I use the fact that $CD=\dfrac{4}{10}CD$? Thank you in advance! Any help would be appreciated.

I have not studied trigonometry.

Answer 1

Draw the radius from $O$ to $AC$ at $P$. Triangles $ADC$ and $OPC$ are similar and $OP = OD$. Enough?

EDIT: Let the height $CD=h.$ Then $OD = .4h$ and so $CO = .6h.$ You have

$$\frac{2}{3} = \frac{.4h}{.6h} = \frac{OP}{OC} = \frac{AD}{AC} =\frac{c}{2a}.$$

Now use that the perimeter is $40.$

Answer 2

Let $|AB|=c$, $|CD|=h_c$, $|OD|=r=\tfrac2{5}\,h_c$, $\rho=\tfrac12\,P=20$. Then using the formula for the area, \begin{align} S_{ABC}&=\tfrac12\,c\,h_c ,\\ c&= \frac{2\,S_{ABC}}{h_c} = \frac{2\rho\,r}{h_c} = \frac{2\rho\,\tfrac25\,h_c}{h_c} = \tfrac45\rho =16 . \end{align}