Isosceles triangle and inscribed circle

Question

If the center of the inscribed circle in a isosceles triangle is dividing the height of the triangle in a proportion 5 : 12, what's the lenght of the legs if the base is 50cm?

Answer 1

Let us draw a picture for the story:

Here $AB$ is the base with the mid point $O$ (placed in the origin of the axes in the picture), and $C$ is on the side bisector of $AB$. The triangle from the $OP$ is $\Delta ABC$, and we know that it has the incenter $I$ delimiting the proportion $5:12$ on the height $CO$. First of all, the "samller part" is $IO$, because $O$ is the biggest angle in $\Delta CAO$. So we have explicitly, also using the angle bisector theorem in this triangle: $$ \frac{5}{12} = \frac{IO}{IC} = \frac{AO}{AC} = \frac{AB/2}{AC} = \frac{25}{AC} \ . $$ This gives $AC=60$.

$\square$

The picture was drawn using scale one to five.

Answer 2

If we let $x$ be the proportionality constant, then it follows that the height of the triangle is $5x+12x=17x$ and the inradius is $5x$. The area of the triangle, $$A=\frac 12 \times 50 \times 17x$$ Calculated another way, $$A=rs =5x\times \frac{2l+50}{2}$$ Equating the two expressions gives $l=60$ cm.