If two sides of a triangle are equal, and the angle between them is $60^\circ$, prove the third side is equal to the first two sides.

Question

In other words, given points $A$ and $X$. Rotate $X$ $\,-60^\circ$ around $A$ to get point $X'$. How would you prove $XX' = AX = AX'$?

I know this is true.

Answer 1

The base angles of an isosceles triangle are equal, and all three angles must add up to $180^\circ$. $2\theta+60=180\implies\theta=60$, so the triangle is equiangular, and is thus equilateral.

Some other ways:

By cosine rule, $$a^2=b^2+c^2-2bc\cos A=2b^2-2b^2\cos60^\circ=b^2$$ so $|a|=|b|=|c|$, where $a,b,c$ are the three sides.

The area of an equilateral triangle is $s^2\frac{\sqrt 3}4$, where $s$ is the side length. The area of any triangle is $\frac12ab\sin\theta$, where $\theta$ is the angle between sides $a$ and $b$. $$\frac12bc\sin60^\circ=\frac12b^2\frac{\sqrt 3}{2}=b^2\frac{\sqrt 3}{4}=c^2\frac{\sqrt 3}{4}$$

This lacks the last side to prove it equilateral, but it's good verification.

Answer 2

We know we have an isosceles triangle, so all three angles equal $60^\circ$. Applying the Law Of Sines:

$$ \frac{\sin(60^\circ)}{AX}=\frac{\sin(60^\circ)}{XX'} $$

Cross multiply

$$ (XX')\sin(60^\circ)=(AX)\sin(60^\circ) $$

and cancel

$$ XX'=AX $$

Answer 3

The bisector of the $60^\circ$ angle is perpendicular to the side (of length say $r$) opposite the angle because the two sides forming the angle have the same length (say $s$). This bisector splits the triangle into two right triangles, bisects the angle into two $30^\circ$ angles, and bisects the third side into two sides each of length $\frac{r}{2}$. But then $\frac12 = \sin 30^\circ = \frac{\text{opp}}{\text{hyp}} = \frac{r/2}{s} = \frac12\cdot \frac{r}{s}$ so that $r=s$.