What are the angles of this triangle? In triangle ABC, angle A is half of angle C and side BC is half of side AC.
What are the angles?

Sorry about the drawing, lol
I realize that it should be 90°, 60°, 30° but since it’s not given that it’s a right triangle I just can’t seem to arrive to the answer. 
I’m supposed to solve this with only the basic triangle and parallel line theorems, without any trigonometry.
 A: Let $P$ be the midpoint of $AC$.
Draw $BP$, draw a perpendicular from $P$ onto $AB$ (to a point $Q$ on $AB$), and draw a perpendicular from $C$ onto $BP$ (to a point $R$ on $BP$).

This splits it up into four triangles. It is easy to see that $BCR$ and $PCR$ are congruent because they are right-angled triangles with a shared side and the same length hypotenuse ($|CP|=|BC|=x$). Therefore the angles at $C$ are the same, and both $\alpha$.
These two triangles $BCR$ and $PCR$ are also congruent to triangle $PAQ$, because they have their angles and a side length (hypotenuse) in common.  
This means that the inner triangle $BPQ$ is a right triangle, with its hypotenuse twice the length of one side ($|BP|=|BR|+|RP|=2|PQ|$), i.e. a 30-60-90 triangle.
Looking at the angles that meet at $P$, we see that $60+2(90-\alpha)=180$, so $\alpha=30$.
A: Call $u=AB$. Let the bisector of $2\alpha$ meet $AB$ in $D$. We know that $DA=\frac23u$ and from triangle $ADC$, which is isosceles, that the length of the bisector $CD$ is  $\frac23u$. 
The bisector of $\angle ADC$ meets $AC$ in $E$.  Again, as $\triangle ADC$ is 
isosceles, we have that $\angle DEC$ is $\pi/2$.  
Finally $\triangle DEC$ is congruent to $\triangle DBC$ since $CB=CE$, $CD=CD$ and $\angle ECD=\angle BCD$, hence $\angle ABC=\pi/2$.
A: I got $$\frac{3\gamma}{2}+\beta=\pi$$ so
$$\beta=\pi-\frac{3\gamma}{2}$$
and by the theorem of sines we get
$$\frac{\sin(\pi-\frac{3}{2}\gamma)}{\sin(\frac{\gamma}{2})}=\frac{b}{\frac{b}{2}}=2$$
so we have to solve
$$\sin(\frac{3\gamma}{2})=2\sin(\frac{\gamma}{2})$$
A: 
Draw the line from A to meet the extended CB at D such that AB bisects ∠DAC and let E be the midpoint of AC. 
Since ∠DAC = ∠ACD = $2\alpha$ and AE = EC, △ADE and △CDE are congruent, with ∠DEC = 90 and DE bisects ∠ADC. Since AB and DE are both angles bisectors, the point P is the incenter of △ADC and CP bisects ∠ACD. 
Since ∠ECP = ∠BCP and CE = CB, △CEP and △CBP are congruent, which leads to ∠ABC = ∠DEC = 90.
Thus, $\alpha = \frac{90}{3} = 30$.
A: (This answer is unintentionally similar to @MichaelHoppe's.)


$$\underbrace{\overline{AD}\cong\overline{CD}}_\text{Isos $\triangle$ Thm}\quad\to\quad
\underbrace{\triangle CBD\cong \triangle CMD}_\text{SAS}\quad\to\quad \angle B = 90^\circ$$
