Can we always find a point $F$ in the line $BC$ of any triangle such as above so that the triangle $APQ$ is an isosceles triangle and the area of the triangle $AreaAPQ=AreaABC$?
It'll be grateful if someone could help me with this.
Given $AF$ is the angle bisector
On
hint....Since the area of the triangle is $$\frac 12bc\sin A=\frac 12|AP|^2\sin A$$ you have that $|AP|=|AQ|=\sqrt{bc}$
It therefore boils down to constructing the geometric mean of the two lengths $AB$ and $AC$ which is given here http://planetmath.org/compassandstraightedgeconstructionofgeometricmean So you can mark off the distances AP and AQ and thus find $F$
It is not always possible.
You need equality of areas of triangles $\triangle FQB$ and $\triangle FPC$. Let $\theta$ the angle $\angle QFB=\angle PFC$. Since $\overline{FQ}=\overline{FP}$ we have $$\frac 12 \overline{FB}\cdot\overline{FQ}\sin \theta=\frac 12 \overline{FC}\cdot\overline{FP}\sin \theta\iff\overline{FB}=\overline{FC}$$ which is not always the case.