No, but the diagonals of a square does bisects its interior angles.
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Only a square and a rhombus will have all its diagonals bisecting vertices. In other shapes some - but not all - diagonals can bisect vertices.
Suppose that the parallelogram is a rhombus (a parallelogram with equal sides). If we draw the diagonals, isosceles triangles are formed (where the median is also an angle bisector and perpendicular to the base). Since the diagonals of a parallelogram bisect each other, and the diagonals don't bisect the vertex angles where they are drawn, then the parallelogram is not a rhombus.
Since the diagonals of a rhombus are perpendicular between them, then in one forth part of the rhombus they form a right triangle where hypotenuse is the side of the rhombus, the base and the height are one half part of its diagonals. Let's take a look at this right triangle.The base and the height lengths could be congruent if and only if the angles opposite to them have a measure of 45⁰, which is impossible to a rhombus because these angles have different measures as they are one half of the two adjacent angles of the rhombus (the diagonals of a rhombus bisect the vertex angles from where they are drawn), which also have different measures (their sum is 180⁰ ).Therefore, the diagonals of a rhombus are not congruent as their one half are not (the diagonals of a rhombus bisect each other).
Both are quadrilaterals. Both have two pairs of side of equal length. In a kite they are adjacent sides, in a rectangle they are opposite. A kite has one pair of equal angles, all of a rectangle's angles are equal. In a kite, one diagonals bisects the other, in a rectangle both do.
Not always. 1. The median to the base of an isosceles triangle bisects the vertex angle. 2. When the triangle is an equilateral triangle, then the medians bisect the interior angles of the triangle.