# Angle Bisectors and Perpendiculars in a QuadrilateralWhat is this about? A Mathematical Droodle

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The angle bisectors of a quadrilateral ABCD form a cyclic quadrilateral XYZW. The applet suggests that the perpendiculars from the vertices of the latter to the corresponding sides of the former form a third quadrilateral, say A1B1C1D1, this one inscriptible! [Grinberg] The result is true for both external and internal angle bisectors. However, for technical reasons, it is preferable to work with the external case.

In the applet, X is the intersection of the bisectors at A and B, Y is the intersection of the bisectors at B and C, etc., X' is the foot of perpendicular from X to AB, etc. Also, M denotes the center of the circle circumscribed around XYZW and X1 the foot of perpendicular from M to XX', etc. We wish to show that

 (1) MX1 = MY1 = MZ1 = MW1.

(In addition to the claim above, (1) asserts that the two circles at hand - one circumscribed around XYZW, the other inscribed in A1B1C1D1 - share the center.)

Since ZW is the exteranl bisector of angle at D, in the right triangles DWW' and DZZ' the angles at D are equal, and so are the other pair of angles:

∠DWW' = ∠DZZ',

which is the same as

 (2) ∠ZWW1 = ∠WZZ1.

On the other hand, since M is the center of the circle circumscribed around XYZW,

 (3) MZ = MW.

If so, ΔZMW is isosceles. Its base angles are equal, such that

 (4) ∠MZZ1 = ∠MWW1,

as the differences of equal angles. We now have two right triangles MZZ1 and MWW1 with equal angles and the hypotenuses, see (3). Their corresponding sides are equal, and the last equality in (1) follows. The others are obtained by a cyclic permutation of vertices.

### Remark

The statement we just established has an interesting special case where the given quadrilateral ABCD is inscriptible. In this case, the perpendicular bisectors of the sides ABCD, too, form an inscriptible quadrilateral.

### Reference

1. D. Grinberg, A Tour Around Quadrilateral Geometry