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Construct, with ruler and compass, a triangle $ABC$ knowing the angle $\widehat{A}$ and $m_a$ and $m_b$, where $m_a$ and $m_b$ are the medians relative to the vertices $A$ and $B$, respectively.

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  • $\begingroup$ It looks to be an easy one, but I wasn't able to do it... $\endgroup$ – Ders Nov 2 '14 at 11:50
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If we set $m_a=3x,m_b=3y$ and consider the following construction: enter image description here

We have that: $$GU = 2x\sin\varphi,\quad GV=x\sin\varphi,\quad UV=M_BW=3x\sin\varphi,$$ $$M_B V^2 = y^2-x^2\sin^2\varphi,\quad AU=2x\cos\varphi,\quad AW=AU-M_BV, $$ hence: $$\cot A = \frac{AW}{M_B W}=\frac{2x\cos\varphi-\sqrt{y^2-x^2\sin^2\varphi}}{3x\sin\varphi}$$ and we have: $$3\cot A = 2\cot\varphi-\sqrt{-1+\frac{y^2}{x^2}(1+\cot^2\varphi)},\tag{1}$$ so it is possible to get the value of $\cot\varphi$ (then build the triangle $GAB$ and the triangle $ABC$) by solving the quadratic equation: $$\left(4-\frac{y^2}{x^2}\right)\cot^2\varphi - 12\cot A\cot\varphi +\left(9\cot^2 A+1-\frac{y^2}{x^2}\right)=0.\tag{2}$$

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