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
 A: 
As in the attached diagram, let $ABC$ be the original equilateral triangle and let $D$ be a point in $\triangle ABC$. 
We let point $E$ be on the opposite side of $BC$ as $D$ such that $\triangle BDE$ is equilateral. Then $BD=BE$, $BA=BC$ and $\angle DBA=\angle EBC=60^{\circ}-\angle DBC$. And therefore $\triangle DBA$ and $\triangle EBC$ are congruent. This implies that $EC=DA$ and since $DE=BD$, we now have $\triangle CDE$ as the triangle we want.
Let $\angle ADB=x$ and $\angle BDC=y$. Then $\angle EDC=y-60^{\circ}$, $\angle DEC=x-60^{\circ}$ and $\angle DCE=300^{\circ}-x-y$ are our desired angles.
A: 
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.
