Euclidean geometry problem without trigonometry

Question

$ABC$ is a triangle such that the median divides the angle $A$ in $15$ e $30$ degress. What the measure of the other two angles? I tried building a parallelogram on the MC side but I didn't know nothing of useful. The answer is $30$ degrees one and $105$ other

Answer 1

Reflect point B over $\overline{AD}$ such that the reflected point $B$ be $B'$ now draw line segments as shown in the above diagram.

Let O be the circumcenter of $\triangle{BAC}$

As $BD=DC=DB'$ by simple angle chasing you can find out that $\angle BB'C = 90$

As B' is a reflection of B $\triangle ABB'$ is an isosceles triangle and $\angle BAB'=30,\angle BB'A=75=\angle ABB'$. Therefore $AB'C=15$ hence $AC=CB'$

Now refer to the $\triangle BOB'$ it is equilateral because O is the circumcenter of the $\triangle BAB'$

By a very simple proofing u can get that $\triangle AOB' \cong \triangle ACB'$ and $AO=OB'=AC=CB'$

Since $\triangle BOB'$ is equilteral $BB'=B'C$

We already know that in $\triangle BB'C=90$ and by the previous relationship you can find out the triangle is a 45-90-45 one.

Let segments marked in '|' be equal to $a$ because $\triangle BDB$ is 45-90-45 lines marked in '||' will be equal to $a\sqrt{2}$

Now draw line $CI \perp BA$

By simple angle chasing u can get $\triangle AIC$ is a 45-90-45 triangle and $\angle ACI=45$ because $AC=a\sqrt{2}, IC=a$

Now look at the $\triangle BIC$, $IC=a$ and $BC=2a$

Then simply u can get $\angle ICB=60, IBC=30$

Then the final answer will be $\angle ABC=30$ and $\angle ACB=45+60=105$

Answer 2

Let $M$ be the midpoint of $\overline{BC}$. Taking $\angle BAM$ to be the $15^\circ$ angle at $A$, drop perpendiculars from $B$ and $M$ to $B'$ and $M'$ on an extension of $\overline{AC}$.

By the $30^\circ$-$60^\circ$-$90^\circ$-ness of $\triangle AMM'$, and the $1:2$ similarity of $\triangle CMM'$ and $\triangle CBB'$, we can write $$|AM|=2|MM'|=|BB'|$$ Since $\triangle ABB'$ is an isosceles right triangle, the above implies that $|AM|=|AB'|$, so that $\triangle AMB'$ is isosceles with vertex angle $30^\circ$. Noting also that $\overline{MB'}$ divides $\triangle CBB'$ into two isosceles triangles, a short angle-chase gives the result. $\square$

Answer 3

In the figure, $E$ is the reflection of $C$ about $AM$.

In other words, $AM$ is the perpendicular bisector of $CE$.

(1) $\angle CAM = 30^o \implies \angle CAE=60^o$

(2) $\because AC=AM, \Delta ACE$ is an equilateral triangle.

(3) Thus $AE=EC $

(4) $\because AM$ perpendicularly bisectors $CE$, $ EM=MC$

(5) Since $BM=MC=EM$, $EB \text{ }\bot \text { }EC$

(6) Note that $AM \text{ }\bot \text { }EC$

(7) Thus $AM // EB$

(8) Thus $\angle EBA=\angle MAB =15^o$

(9) $\because \angle BAE=15^o, \therefore \angle EBA= \angle BAE$

(10) $\therefore AE=EB $

(11) Hence $EC=EB$ and $\Delta CEB$ is an isosceles right angled triangle.

(12) $\therefore \angle ABC= 45^o-15^o=30^o$

(13) $\angle ACB=60^o+45^o=105^o$

Answer 4

Note $m=\tan(15^{\circ})=\sqrt{\dfrac{1-\cos(30^{\circ})}{1+\cos(30^{\circ})}}=\sqrt{\dfrac{2-\sqrt3}{2+\sqrt3}}$ and let the lines $AM: y=mx$ and $AC: y=x$ in the triangle $\triangle{ABC}$ below. All parallel to the side $ {BC}$ makes the same angle with the $x-$coordinate. So, we can choose for the side $AC$ a length $\sqrt2$ so $AC=CC’=1$. The problem is to determine the length $C’B=a$ such that $AM$ be a median of the triangle.

We have the system $$AM: y=mx\\ AC: \frac{y-1}{x-1}=-\frac 1a$$ from which $x=\dfrac{a+1}{am+1}$. It follows because $M$ is the midpoint of side $BC$ that $$\dfrac{a+1}{am+1}=\frac{2+a}{2}$$ so $$a=\dfrac{1-2m}{m}=\frac{1-2\sqrt{\dfrac{2-\sqrt3}{2+\sqrt3}}}{\sqrt{\dfrac{2-\sqrt3}{2+\sqrt3}}}=\frac{\sqrt{2+\sqrt3}-2\sqrt{2-\sqrt3}}{\sqrt{2-\sqrt3}}$$ Hence, we have a right triangle of sides $1,a, \sqrt{1+a^2}$.

Note that determining this way the angle $\angle{ABC}$ without trigonometry would be impossible if the relation $\dfrac1a$ of the legs did not give the tangent of some known angle. In our case we have $$\dfrac{1}{\dfrac{\sqrt{2+\sqrt3}-2\sqrt{2-\sqrt3}}{\sqrt{2-\sqrt3}}}= \frac{\sqrt{2-\sqrt3}}{\sqrt{2+\sqrt3}-2\sqrt{2-\sqrt3}}$$ which reduces without difficulty to $\dfrac{1}{\sqrt3}$ which corresponds to an angle of $30^{\circ}$. We are done.