Random point inside an equilateral triangle Take any equilateral triangle and pick a random point inside the triangle.
Draw from each vertex a line to the random point. Two of the three angles at the point are known let's say $x$,$y$.
If the three line segments from each vertex to the random point were removed out of the original triangle to form a new triangle , what would the new triangle's angles be?

Video about the problem


In the attached figure, EC, EA and AG are parallel to and of equal length as AD, CD and DB respectively. The angles opposite to them are $120^{\circ}$ since $\triangle EFG$ is exactly inverted to $\triangle ABC$. Let's call these points I, J and K. Hence EI = EJ, AJ = AK and CI = GK due to the same reason. Thus the green triangles can be perfectly joined to form the red triangle and the angles will be $60^{\circ}$ less than the given angles i.e. $x-60^{\circ}$, $y-60^{\circ}$ and $300^{\circ}-x-y$ since EADC and the rest (not shown in diagram for simplicity) are parallelepipeds.
