reverse proof of sin30 = 1/2 without trigonometry

Question

The length of one side of the right triangle (AB) in the picture is half of the length of the hypotenuse (AC).

How can I prove that $\angle ACB = 30^0$ , without using trigonometry?

Answer 1

Extend a point $D$ so that $D$ is colinear to $AB$ and $BD = AB$.

So $\triangle ABC \cong DBC$ by Side ($AB = BD$) Angle ($\angle ABC \cong \angle DBC$ are both right angles) Side $BC=BC$ so $AC=DC$ and $AD = AB+BD = \frac 12 AC + \frac 12 AC= AC$.

So $\triangle ADC$ is equilateral and $\angle ACB=\frac 12 ACD = 60$. So $\angle ACB = 30$.

===old ====

Then triangle $ABC$ will have angles $60, 60, 60$ so it is an equilateral triangle. $AD = AC$ and $AB = \frac 12 AC$.

So by Pyth Theorem $BC^2 + AB^2 = AC^2$ so $BC^2 + (\frac 12 AC)^2 = AC^2$.

Answer 2

Extend line $AB$ to point $X$ such that $AB$ = $BX$. $AXC$ is an equilateral triangle as $AX = AB + BX = 2AB$. $CB$ is perpendicular to $AX$ so it is the median, so $\angle ACB = \angle {ACX\over2} = \frac {60\deg}2 = 30\deg$