Consider convex quadrilateral $ABCD$. Let there be a point $P$ in the interior of the quadrilateral such that $PA = PB$ and $PC = PD$.
$K,L, M$ are the mid-points of $AB , BC , CD$ respectively. Prove that $\angle APB = 120^{\circ}$ if triangle $KLM$ is equilateral.
In Fig. 1, we let N be the constructed midpoint of AD.
By midpoint theorem, $2KL = AC= 2MN\tag 1$
Similarly, $2NK = BD = 2ML\tag 2$
Therefore, $KLMN$ is a parallelogram $\tag 3$
Because $\triangle KLM$ is equilateral, $KL = LM = MK\tag 4$
[(3), (4)] $\implies KLMN$ is a rhombus$\tag 5$
[(1), (2), (5)] $\implies AC = BD\tag 6$
Also, [(4), (5) $\implies \angle NKL = 120^\circ\tag 7$
In Figure 2, we add in the lines $PA$, $PB$, $PC$, $PD$. And, let $AC$ intersect $BD$ at $Q$.
In $\triangle PAC$ and $\triangle PBD$, $PA = PB$ (given), $PC = PD$ (given), $AC = BD$ (proved in (6))
Therefore, $\triangle PAC$ and $\triangle PBD$ are congruent.
Then, $\angle PAC = \angle PBD$
In Fig. 3, from the above, $P$, $A$, $B$, $Q$ are con-cyclic.
By angles in the same segment, $\angle APB = \angle AQB$
But $\angle AQB$ can be proved to be equal to $\angle NKL$ (through a series of parallel lines).
By (7), $\angle APB = 120^\circ$