Consider a triangle $ABC$ where $\measuredangle A=60^{\circ}$. Draw its bisection intersecting $BC$ at $D$. Let $AB = x$, $BD = y$ and $AC=x+y$, $\measuredangle ABC = \alpha$ and $\measuredangle ACB = \beta$. Find $\alpha$ and $\beta$.

Answer: $\alpha = 80^{\circ}$ and $\beta = 40^{\circ}$.

The problem came with a picture, which I've redone in paint (as it was handdrawn to me).

enter image description here

I've tried a couple of things. First I marked a point $E$ in $AC$ such that $AE = x$ and drew $BE$. This created an equilateral triangle $AEB$ and another triangle $BEC$ with angles $\measuredangle BEC = 120^{\circ}$, $\measuredangle EBC = \alpha - 60^{\circ}$ and $\measuredangle ECB = \beta$. I couldn't go anywhere from here.

Next I've extended $AB$ to a point $E$ such that $AE = x+y$ and closed a triangle $AEC$. This triangle is again equilateral, and creates another triangle $BEC$ with angles $60^{\circ}$, $60^{\circ} - \beta$ and $60^{\circ}+\beta$. This too hasn't helped much.

I've established that if we can prove that $\alpha - \beta = 40^{\circ}$ we can find $\alpha$ and $\beta$, since $\alpha+\beta = 120^{\circ}$ by the original triangle.

  • $\begingroup$ Could $y/x=DC/(x+y)$, then $BC$ and then sine or cosine theorem work? $\endgroup$ – Alexey Burdin Apr 24 '15 at 12:04
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    $\begingroup$ @MichaelRozenberg Thanks. I didn't know such a symbol existed. It looks much nicer! $\endgroup$ – Mark Fantini Mar 24 '17 at 13:37

Draw $DE$ where $E$ lies on $AB$ at a length of $x$ units from $A$ and $y$ units from B. Then, triangles $ABD$ and $AED$ will be congruent. (2 equal sides and the angle between them is also equal). Now, consider the triangle $DEC$. The sides $EC$ and $ED$ are equal (both equal to $y$). Thus, angles $EDC$ and $ECD$ will be equal. $\measuredangle EDC = 2\alpha - 120^{\circ}$ and $\measuredangle ECD = \beta$. This gives us $2\alpha - 120^{\circ} = \beta$. This, along with $\alpha + \beta = 120^{\circ}$ gives us $\alpha = 80^{\circ}$ and $\beta = 40^{\circ}$.


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