Notes
three triangles in a circle in a triangle solution

Solution to the Three Triangles in a Circle in a Triangle Puzzle

Solution by Angles in the Same Segment, the Cosine Rule, and the Intersecting Chords Theorem.

In the above diagram, the point $O$ is the centre of the circle, the point $F$ lies on the circumference so that triangle $B F D$ is isosceles, and $E$ is such that $E G C$ is a straight line.

Since triangle $B G A$ is equilateral angle $B \hat{A} G$ is $60^\circ$ and so, using the result that angles in the same segment are equal, angle $B \hat{F} D$ is $60^\circ$ and so since triangle $B F D$ is, by construction, isosceles it is in fact equilateral. Its area is unaffected by rotating it, and it can be rotated to bring $F$ to where the circle touches one side of the outer equilateral triangle. This brings the configuration to the following diagram in which it is clear that the area of triangle $B F D$ is one quarter of the outer triangle.

Let $a$ be the length of $A B$, $b$ of $G C$, $c$ of $G D$, $d$ of $B D$. Angle $B \hat{G} D$ is $120^\circ$ since both $B \hat{G} C$ and $C \hat{G} D$ are $60^\circ$. Applying the cosine rule to triangle $B G D$ yields the identity:

Using $\cos(120^\circ) = - \frac{1}{2}$, this simplifies to:

As $A G$ is a chord of the circle, the perpendicular bisector of $A B$ passes through $O$. Since triangle $A G B$ is equilateral, that bisector also passes through $G$ and is a line of symmetry of triangle $A G B$. Angle $O \hat{G} C$ is therefore $90^\circ$ and so under reflection through the line extending $O G$ the point $C$ is taken to $E$ meaning that the length of $G E$ is the same as that of $G C$, namely $b$.

Applying the intersecting chords theorem to the chords $A D$ and $E G$ then shows that $a c = b^2$. Putting this in the above equation gives:

Multiplying through by $\frac{\sqrt{3}}{4}$ shows that the areas of the three shaded triangles is the same as that of triangle $B F D$ and hence is a quarter of the area of the outer triangle.