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Given two angles of a triangle, finding an angle formed by a median

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Problem: Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median of $\triangle ABC$. Show that $\angle MBC=\angle BAC$. Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median of $\triangle ABC$. Show that $\angle MBC=\angle BAC.

$\def\vecl{\overrightarrow}$

Remark: One may of course try to use congruent or similar triangles. I am wondering if one can solve the problem by using vectors in $\mathbf{R}^2$: for instance, can one somehow show by manipulating vectors that $$ \frac{\vecl{BM}\cdot\vecl{BC}}{|\vecl{BM}||\vecl{BC}|}=\cos(\frac{\pi}{6})\ ? $$

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I think you mean $\frac\pi6$ instead of $\frac\pi3$. (2 comments)

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This is a synthetic geometry approach to the problem. It turned out more involved than I expected, and yet I haven’t found anything that could be omitted! I don’t say there isn’t a shorter way to do it, though, probably using some theorems that I haven’t thought of. (Maybe circle inversion? Seems like someone always pulls out circle inversion for knotty geometry problems.)

I also have a nice, direct solution using trigonometry. I haven’t yet solved it using vectors as you asked, because in the middle of my vector approach I keep reaching for trigonometric relationships, like the sine rule.


So here’s the situation as it’s given.

Triangle ABC, with angle A = 30° and angle C = 15°. The midpoint M of the side AC is marked, and the median BM drawn.

To begin, double the known angles: reflect $AC$ across $AB$ to double the 30° angle, and reflect $AC$ across $BC$ to double the 15° angle. Call the intersection of these reflected lines $D$. Since our newly doubled angles are $\angle CAD = 60°$ and $\angle ACD = 30°$, the angle sum of $\triangle ACD$ tells us that $\angle ADC = 90°$.

The previous diagram with point D added, with lines connecting it to A and C forming a 90° angle at D.

In a right triangle like $\triangle ADC$, the circumcentre is the midpoint of the hypotenuse—$M$ in this case. So $|AM| = |DM|$, and $\triangle ADM$ is at least isosceles. But since $\angle MAD = 60°$, this triangle is in fact equilateral, with $|AM| = |DM| = |AD|$.

The previous diagram with most of a circle visible, centred at M and passing through A, D, and C. Triangle ADM is shown to be equilateral.

Now, since equilateral triangles are so charmingly symmetrical, the angle bisector $AB$ is also the perpendicular bisector of the opposite side $DM$ (or would be, if we extended it). That makes any point on it, $B$ included, equidistant from $D$ and $M$. So $\triangle BDM$ is isosceles.

The previous diagram with line segment BD added, marked as being the same length as BM.

Speaking of angle bisectors, the angle bisectors of any triangle concur at its incentre. As the intersection of $AB$ (bisecting $\angle CAD$) and $BC$ (bisecting $\angle ACD$), $B$ is the incentre of $\triangle ACD$. As $BD$ connects this incentre to the remaining vertex, it is also an angle bisector, bisecting $\angle ADC$. As that’s a right angle, $\angle BDC = 45°$.

Let $X$ and $Y$ be the tangent points of the incircle on sides $CD$ and $AC$, respectively. As radii and tangents are perpendicular, $BX \perp CD$ and $BY \perp AC$.

The previous diagram with the outer circle removed and the incircle drawn. Tangent points X, Y and Z are drawn and the radii touching them marked as equal lengths, and angle BDC is marked as 45°.

By the RHS criterion, $\triangle BMY \cong \triangle BDX$, so $\angle BMY = 45°$ as well.

The previous diagram with the incircle removed. Angle BMY is marked as 45°.

Hence, its complementary angle $\angle BMC = 135°$, and by the angle sum of $\triangle BCM$, $\angle CBM = 30°$. ∎

The previous diagram with everything removed except the original parts and the 45° angle BMA. Angle BMC is marked as 135°, and angle BCM is marked as 30°.

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