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 $$ \frac{\vecl{BM}\cdot\vecl{BC}}{|\vecl{BM}||\vecl{BC}|}=\cos(\frac{\pi}{6})\ ? $$

euclidean-geometry vector

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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 $\triangle BCM$, $\angle CBM = 30°$. ∎

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