# Two Squares and an Equilateral Triangle +-- {.image} [[TwoSquaresandanEquilateralTriangle.png:pic]] > Two squares and an equilateral triangle. What’s the angle? =-- ## Solution by [[Equations of Lines]] +-- {.image} [[TwoSquaresandanEquilateralTriangleLabelled.png:pic]] =-- Position the diagram on a coordinate grid so that point $A$ is at the origin and point $C$ at the point $(1,0)$. From the properties of [[equilateral triangles]], the coordinates of $G$ can be seen to be $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and so the equation of the line extending $C G$ is: $$ y = -\sqrt{3} x + \sqrt{3} $$ Point $F$ lies on this line and also on the line $y = x$. Its coordinates can therefore be calculated to be $\left( \frac{\sqrt{3}}{1 + \sqrt{3}}, \frac{\sqrt{3}}{1 + \sqrt{3}}\right)$. The length of $B C$ is then $1 - \frac{\sqrt{3}}{1 + \sqrt{3}} = \frac{1}{1 + \sqrt{3}}$ and so the coordinates of $D$ are $\left(1, \frac{1}{1 + \sqrt{3}}\right)$. The length of $G H$ is therefore: $$ \frac{\sqrt{3}}{2} - \frac{1}{1 + \sqrt{3}} = \frac{ \sqrt{3}(1 + \sqrt{3}) - 2}{2(1 + \sqrt{3})} = \frac{\sqrt{3} + 1}{2(1 + \sqrt{3})} = \frac{1}{2} $$ The triangle $G H D$ is therefore an [[isosceles]] [[right-angled triangle]] and so angle $G \hat{D} H$ is $45^\circ$.