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Explanation

### Explanation

The applet illustrates a beautiful discovery of Professor Duane DeTemple of Washington State University.

Let ABCD be a convex quadrilateral. The successive exterior angle bisectors of ABCD intersect pairwise in points E, F, G, H. Then EFGH is cyclic. But that's not all. From each of the points E, F, G, H drop a perpendicular to the corresponding side of ABCD and let the extended lines meet in successive pairs in P, Q, R, S. Then PQRS is an inscriptible quadrilateral and, moreover, its incircle is concentric with the circumcircle of EFGH.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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The exterior angles of a polygon are supplementary to its interior angles. So we see that, for example,

 ∠ABE = (180° - ∠ABC)/2 and ∠BAE = (180° - ∠BAD)/2,

from which

 ∠FEH = ∠AEB = 180° - ∠ABE - ∠BAE = (∠ABC + ∠BAD)/2.

Similarly,

Adding the two and bearing in mind that the sum of the angles of a quadrilateral equals 360°, we obtain

 ∠FEH + ∠FGH = 180°,

which shows that EFGH is indeed a cyclic quadrilateral.

Consider now ΔEHS:

 ∠SEH = ∠SEA = 90° - ∠BAE = 90° - (180° - ∠BAD)/2 = ∠BAD/2.

In exactly same manner, also ∠EHS = ∠BAD/2. It follows that ΔEHS is isosceles, so that the angle bisector at S is also the perpendicular bisector of EH. Similar considerations show that the remaining angles bisectors in PQRS are the perpendicular bisectors of the remaining sides of EFGH. But the latter being cyclic, the perpendicular bisectors of its sides meet in the center of its circumcenter. Applied to PQRS, this means that its angle bisectors meet in the same point, which is thus equidistant from the sides of PQRS, proving the claim.

Obviously, at times PQRS degenerates to a point, in which case, the point is simply the center O of the circumcircle of EFGH. As regard the latter, Professor DeTemple also observes that

1. EFGH is always cyclic;
2. When ABCD is a parallelogram EFGH is a rectangle;
3. When ABCD is a rectangle, EFGH is a square.

### References

1. R. Honsberger, From Erdös To Kiev, MAA, 1996, pp. 63-65.