<s>[INST] Two points on sides AB and AC of a triangle How to determine using only the straightedge and compass the points P and Q on the sides AB and AC of a given triangle ABC such that the triangle APQ and the quadrilateral BPQC have the same surface and the segment PQ has the minimal possible length?
 [/INST] The least $PQ$ occurs when $AP = AQ$.
Let $AB=c$, $AC=b$ and $m(\angle A)=\alpha$.
We know that $Area_{APQ} = \frac{1}{2} Area_{ABC}$, then we have:
$$\frac{1}{2} AP^2\sin\alpha =\frac{1}{2} \frac{1}{2} cb\sin\alpha \Rightarrow$$
$$AP=\sqrt{\frac{cb}{2}}.$$
Therefore $AP$ is the geometric mean of $c$ and $\frac{b}{2}$.
We only have a solution if $\frac{c}{2} < b <2c$.
The construction is shown below:

EDIT.
(A short proof of the first statement)
Let $Area_{ABC}=S$, and $AP =x$ (where $S$ and $x$ are positive real numbers), then:
$$\frac{1}{2} AP \cdot AQ \sin\alpha = \frac{S}{2} \Rightarrow$$
$$\Rightarrow x AQ= \frac{S}{\sin\alpha} \Rightarrow$$
$$\Rightarrow AQ= \frac{S}{x \sin\alpha}. \quad (1)$$
But we know that
$$PQ^2=AP^2 + AQ^2 -2AP \cdot PQ \cos \alpha \quad(2) $$
Substituting $(1)$ in $(2)$ we get:
$$PQ^2= x^2 + (\frac{S}{x \sin\alpha})^2 - \frac{2S}{\tan \alpha} \Rightarrow $$
$$PQ= \sqrt{x^2 + (\frac{S}{x \sin\alpha})^2 - \frac{2S}{\tan \alpha}}. \quad(3) $$
At the minimal point the derivative of $PQ$ equals zero, so:
$$2x+\frac{-2S^2}{x^3 \sin^2\alpha }=0 \Rightarrow$$
$$\Rightarrow x^4= \frac{S^2}{\sin^2\alpha} \Rightarrow$$
$$\Rightarrow x=\sqrt{\frac{S}{\sin\alpha}}. \quad(4)$$
Substituting $(4)$ in $(1)$, we get:
$$AP=AQ=x=\sqrt{\frac{S}{\sin\alpha}}$$
Therefore the least $PQ$ occurs when $AP=AQ$.
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