Answer:
[tex]18\sqrt{2}[/tex]
Step-by-step explanation:
Consider triangle ABC with the base AC=36. Let the line DE be parallel to the line AC. If DE||AC, then triangles ABC and DBE are similar. If k is the scale factor of these triangles, then
[tex]\dfrac{DE}{AC}=\dfrac{BG}{BF}=k.[/tex]
Thus,
[tex]DE=k\cdot AC=36k,\\ \\BG=k\cdot BF.[/tex]
The area of the triangle ABC is
[tex]A_{ABC}=\dfrac{1}{2}AC\cdot BF=\dfrac{1}{2}\cdot 36\cdot BF=18BF.[/tex]
The area of the triangle DBE is
[tex]A_{DBE}=\dfrac{1}{2}DE\cdot BG=\dfrac{1}{2}\cdot 36k\cdot kBF=18k^2BF.[/tex]
Since line DE divides triangle ABC in two equal area parts, we have that
[tex]A_{DBE}=\dfrac{1}{2}A_{ABC},\\ \\18k^2BF=\dfrac{1}{2}\cdot 18BF,\\ \\k^2=\dfrac{1}{2},\\ \\k=\dfrac{1}{\sqrt{2}}[/tex]
and
[tex]DE=\dfrac{1}{\sqrt{2}}\cdot 36=18\sqrt{2}.[/tex]
