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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.

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  • $\begingroup$ 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. $\endgroup$ – Umberto P. May 14 at 13:57
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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 $40.$

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  • $\begingroup$ 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 $AB$? $\endgroup$ – LYI May 14 at 14:08
  • $\begingroup$ I don't know why I can't tag you. I am sorry. $\endgroup$ – LYI May 14 at 14:16
  • $\begingroup$ See edits. I would automatically get a ping when someone responds to my answer, so there's no need to tag me. $\endgroup$ – B. Goddard May 14 at 14:25
  • $\begingroup$ Oh, okay! I didn't know that! You forgot to add the zeros. $\endgroup$ – LYI May 14 at 14:31
  • $\begingroup$ $\dfrac{AD}{AC}=\dfrac{c}{2a}$. $\endgroup$ – LYI May 14 at 14:47
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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}

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