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An arbitrary quadrilateral and its diagonals. Bases of similar triangles are parallel to the blue diagonal. Ditto for the red diagonal. The base pairs form a parallelogram with half the area of the quadrilateral, A q, as the sum of the areas of the four large triangles, A l is 2 A q (each of the two pairs reconstructs the quadrilateral) while that of the small triangles, A s is a quarter of A ...
The quadrilateral case follows from a simple extension of the Japanese theorem for cyclic quadrilaterals, which shows that a rectangle is formed by the two pairs of incenters corresponding to the two possible triangulations of the quadrilateral. The steps of this theorem require nothing beyond basic constructive Euclidean geometry. [2]
The special case of the theorem for quadrilaterals states that the two pairs of opposite incircles of the theorem above have equal sums of radii. To prove the quadrilateral case, simply construct the parallelogram tangent to the corners of the constructed rectangle, with sides parallel to the diagonals of the quadrilateral.
Proof of the theorem. We need to prove that AF = FD.We will prove that both AF and FD are in fact equal to FM.. To prove that AF = FM, first note that the angles FAM and CBM are equal, because they are inscribed angles that intercept the same arc of the circle (CD).
Labels used in proof concerning complete quadrilateral. It is a well-known theorem that the three midpoints of the diagonals of a complete quadrilateral are collinear. [2] There are several proofs of the result based on areas [2] or wedge products [3] or, as the following proof, on Menelaus's theorem, due to Hillyer and published in 1920. [4]
This is not a cyclic quadrilateral. The equality never holds here, and is unequal in the direction indicated by Ptolemy's inequality. The equation in Ptolemy's theorem is never true with non-cyclic quadrilaterals. Ptolemy's inequality is an extension of this fact, and it is a more general form of Ptolemy's theorem.
Newton's theorem can easily be derived from Anne's theorem considering that in tangential quadrilaterals the combined lengths of opposite sides are equal (Pitot theorem: a + c = b + d). According to Anne's theorem, showing that the combined areas of opposite triangles PAD and PBC and the combined areas of triangles PAB and PCD are equal is ...
The circumcircles of all four triangles of a complete quadrilateral meet at a point M. [7] In the diagram above these are ∆ABF, ∆CDF, ∆ADE and ∆BCE. This result was announced, in two lines, by Jakob Steiner in the 1827/1828 issue of Gergonne's Annales de Mathématiques, [8] but a detailed proof was given by Miquel. [7]