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In a rhombus/parallelogram/trapezoid do the diagonals bisect the corner four angles at the vertices?

Is it possible to construct a non-rhombus by putting four 3-4-5 triangles where the right angle is not at the intersection of the two diagonals?

The following is a parallelogram

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Is there any relationship between the four angles formed at the intersection of the diagonals of a parallelogram and the four corner angles at the vertices of the figure?

Is this true or false, "The four angles formed at the intersection of the diagonals of a parallelogram are congruent." When is it true or false?

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    For parallelograms, see e.g. https://math.stackexchange.com/questions/1811308/in-a-parallelogram-does-the-diagonal-bisect-the-angles-that-they-meet. The diagonals of a parallelogram will bisect its angles if and only if the sides of the parallelogram are all equal length (aka it is a rhombus). – Minus One-Twelfth Oct 28 '23 at 19:46
  • In the trapezoid, note that $\psi = \pi$ as alternate angles of parallel lines. Then $\delta = \psi$ iff the left leg and lower base are equal in length. Similarly $\pi = \Omega$ iff the right leg and upper base are equal in length. – peterwhy Oct 29 '23 at 03:08
  • https://graemewilkin.github.io/Geometry/Quad_Angle_Bisectors.html –  Oct 30 '23 at 12:52

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