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

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.

• You seem to have ignored the fact that the circle centered at $O$ is inscribed in $ABC$. Without that fact there is no way to proceed. – Umberto P. May 14 at 13:57

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 $$40.$$

• Thank you for the response! I appreciate it. Okay! Triangles $ADC$ and $OPC$ are similar because $\measuredangle ADC=\measuredangle CPO=90^\circ$ and they share angle $C$. Therefore, $\dfrac{AD}{OP}=\dfrac{CD}{CP}=\dfrac{AC}{CO}$. How to use this to find $AB$? – LYI May 14 at 14:08
• I don't know why I can't tag you. I am sorry. – LYI May 14 at 14:16
• See edits. I would automatically get a ping when someone responds to my answer, so there's no need to tag me. – B. Goddard May 14 at 14:25
• Oh, okay! I didn't know that! You forgot to add the zeros. – LYI May 14 at 14:31
• $\dfrac{AD}{AC}=\dfrac{c}{2a}$. – LYI May 14 at 14:47

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}