Problem 1. Consider the parallelogram $A B C D$, whose diagonals intersect at $O$. The angle bisectors of $\angle D A C$ and $\angle D B C$ intersect at $T$. It is known that $\overrightarrow{T D}+\overrightarrow{T C}=\overrightarrow{T O}$. Determine the measures of the angles of triangle $A B T$.
Solution.

From the hypothesis, it follows that $D O C T$ is a parallelogram..................................

From $A O \| D T$ we deduce $\angle D T A \equiv \angle O A T \equiv \angle D A T$, so $D A=D T \ldots . \mathbf{2 p}$

Thus $D A=D T=O C$; similarly $B C=C T=O D$, from which $B D=A C$. Thus $A B C D$ is a rectangle, $A O T D$ is a rhombus, triangle $A O D$ is equilateral and triangle $A B T$ is equilateral, so it has angles of $60^{\circ}$ $3 p$