A point $O$ lies inside of the square $ABCD$. Prove that the difference between the sum of angles $OAB, OBC, OCD , ODA$ and $180^{\circ}$ does not exceed $45^{\circ}$.
Source: ToT 2003-JA-5
Tags: inequalities, geometry unsolved, geometry
A point $O$ lies inside of the square $ABCD$. Prove that the difference between the sum of angles $OAB, OBC, OCD , ODA$ and $180^{\circ}$ does not exceed $45^{\circ}$.