[alogo] Collignon, Van Aubel theorem

Construct squares on the sides of a quadrangle ABCD. The segments GE and FH, joining opposite centers of the squares, as shown, are equal and orthogonal to each other.

Orthogonal: Draw the circle (CDG) and (ABE). Extend DL and LC, resp. AJ and BJ, where L resp. J are the intersection points of the circles with GE. Intersecting these lines defines the quadrangle IJKL which is cyclic, because at J and L its angles are right. JL is also a diameter of its circumcircle and the angles at J and L are all 45 deg. Then IK passes through F and H ...

Equal: Split EG = EJ+JL+LG and work with Ptolemy for parts EJ, LG ... Analogously split FH ...
For a simpler proof look at Van_Aubel.html .

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]
[2_0] [2_1] [2_2]

Quadrilaterals with orthogonal diagonals are called "Orthodiagonal". EHGF is an orthodiagonal quadrilateral associated to the arbitrary quadrilateral ABCD. Look at the file Thebault.html for a related figure.

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