find the parallel from the center of the inscribed circle of an isosceles triangle

Question

$ABC$ - isosceles triangle. $O$ - the center of the inscribed circle of the triangle.

$EF$ || $BC$

Given $AB = 24 cm$ and $AF=18cm$, find out $EF =?$

If it's possible, I would be interested in a demonstration without using Fundamental theorem of similarity or the bisector theorem.

Answer 1

$AOF$ and $AHC$ are similar triangles.

$AO=3r$ and $OF=\sqrt{18^2-9r^2}$.

$AOF$ is a right triangle and $$OK=\frac{AO\cdot OF}{AF}$$ So we get the equation $$3r\sqrt{18^2-9r^2}=18r$$ simplify $$\sqrt{324-9r^2}=6$$ square both sides $$324-9r^2=36\to r=4\sqrt{2}$$ So $OF=6$ and $EF=12$.