Circumscriptible quadrilateral

Start with the following property of two trianles t = (AEC) and t' = (CFG), having two angles equal (at C) and two other angles (at E and F correspondingly) complementary. Placing them like in the above figure, we have: (AC/AE) = (CB/EB) = (CG/FG) => (AC/CG) = (AE/FG). Which looks like if we had equal angles at E and F. This proves the following basic property for circumscriptible quadrilaterals (proof idea of A. Varverakis).

The diagonals and the lines joining opposite contact-points, in a circumscriptible to a circle quadrilateral, pass all through a common point O. Indeed, triangles (COH) and (GOD) have equal angles at O and complementary at C and D respectively. Thus (CH/GD) = (OH/OG). Analogous equation will be valid for the ratios of HE and FG. Thus, because CH = HE and GD = GF, CD and EF will meet at a point on the diagonal HG. The same reasoning implies that CD and EF will meet ona point on the diagonal IJ, thus all four lines concure at a point O lying in both diagonals i.e. their intersection point. The proof can also be given by specializing Brianchon's theorem, demonstrated in Brianchon2.html .
The file CircumscriptibleQuadrilateral2.html contains some other aspects of the same subject.

See Also

Brianchon2.html
Circumscriptible.html
Circumscriptible_Construction.html
CircumscriptibleQuadrilateral2.html

 Produced with EucliDraw© http://users.math.uoc.gr/~pamfilos/