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enter image description here 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

Jyotishraj Thoudam
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  • Sure. If $F=C$ the area is less (following your picture, in which $AC<AB$). If $F=B$ the area is greater. Don't see a classical construction instantly..but I expect there is one. – lulu Sep 10 '16 at 14:07
  • That proves that the point $F$ has to lie on the line $BC$ but how would we prove that $F$ must lie on $BC$ and the angle bisector line simutaneously? – Jyotishraj Thoudam Sep 10 '16 at 14:09
  • Maybe I didn't understand what you wanted. I thought you wanted an isosceles triangle with $F$ on the base. You needed $F$ to be the midpoint of the base? – lulu Sep 10 '16 at 14:11
  • Would it be true if ,In other words we prove that $AreaFPC=AreaQFB$? – Jyotishraj Thoudam Sep 10 '16 at 14:11
  • Your problem isn't clear. Are you defining $F$ to be the intersection of the angle bisector and the segment $BC$? That's not an existence question...$F$ clearly exists. Of course you can then ask if $F$ has the property you want. Is that what you meant? – lulu Sep 10 '16 at 14:13
  • Oh I'm sorry, I wanted to know that if $AF$ is the height of the Isosceles triangle satisfying the simultaneity condition. – Jyotishraj Thoudam Sep 10 '16 at 14:14
  • Yes I'm defining $F$ to be the intersection of the angle bisector to $BC$. And I wanted to know if it satifies the simultaneity condition that $AreaAPQ=AreaABC$ – Jyotishraj Thoudam Sep 10 '16 at 14:16
  • It's been so long I'm going over this diorama. I need some input. I'll appreciate it – Jyotishraj Thoudam Sep 10 '16 at 14:20
  • I'm thinking about it. Explicit examples are hard to draw...kind of interesting. I'll post a solution if I find anything worth passing on. – lulu Sep 10 '16 at 14:22
  • Why brands with a slash CF and FB segments? This does mean that its are equal? If AF is the angle bisector it is not true. – Piquito Sep 10 '16 at 14:53
  • Maybe I should change the picture – Jyotishraj Thoudam Sep 10 '16 at 14:54
  • Well, do it. Regards. – Piquito Sep 10 '16 at 14:59
  • Deleted slashs were for segments FQ and FP but it was a dispensable help for readers . – Piquito Sep 10 '16 at 15:10

2 Answers2

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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 $F$

David Quinn
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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.

Piquito
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