Geometry proof question?

The only way a quadrilateral can have two opposite corner angles bisected by a single line is if it is a parallelogram, specifically one with all sides equal (square or rhombus). And line AC is the diagonal. Two adjacent corner angles of the parallelogram always total 180*. If each is bisected, then two halves of adjacent corner angles facing each other must add to 90*. The other diagonal BD meets AC at a right angle because the other two angles of the smaller triangle add up to 90*. That means the third angle of the triangle must be 90*, since the three angles must total 180*.