Proof of construction of $30^{\circ}, 60^{\circ}, 120^{\circ}$ and $135^{\circ}$ angles

Question

What is the proof of constructing $30^{\circ}, 60^{\circ}, 120^{\circ}$ and $135^{\circ}$ angles with ruler and compass? I can prove $90^{\circ}$ by proving that the line joining point of intersection of two circle is perpendicular to the line joining their radius. But I can't prove these angles. Thank You.

Answer 1

For constructing $60^{\circ}$ what we do is, simply construct a equilateral triangle. Since each interior angle of an equilateral triangle is $60^{\circ}$, we're done.

Here $PA=AB=PB$

Rest is just bisecting this angle to get $30^{\circ}$, and reconstruction of this angle on the side $PA$, to double the angle, thus resulting into $120^{\circ}$

What remains is $135^{\circ}$, for this we construct $90^{\circ}$, again $90^{\circ}$ on previous one, and then bisect later one, getting an angle $90^{\circ}+\frac{90^{\circ}}{2}=135^{\circ}$

Answer 2

See the picture above: if we have $60^{\circ}$, it is easy to draw a right angle. When we have this, $135^{\circ}$ is easy because you can draw a square and its diagonal.