Geometry - Problem

[Estoroth]

Geometry

Quadrangle ABCD is written in circle (do You understand this. Written mean
that points A,B,C and D belong to the circle). Prove that centers of circles
written in triangles ABC , BCD, CDA and DAB are the apexes of rectangle.

Bye!
Estoroth

[Bhaskara Aditya]

"Estoroth" <2....@wp.pl.NOSPAM> wrote in message news:<a0voqb$sri$2...@news.tpi.pl>...

Hi!
It's a nice but a pretty simple problem (looks like some national
olympiad or some magazine problem). The figure might look frightening
so draw only the part that interests us. Here is the soln :

(if you like to try the problem, dont read further)

Let ABCD be the cyclic quad.
Let <ADB=<ACB=alpha (denote by a), <ABD=<ACD=beta (denote by b),
<BAC=<BDC=gamma (denote by c), <CAD=<CBD=delta (denote by d).

Note that a+b+c+d = 180 deg.

Let I1, I2, I3 be the incentres of triangles BCD, CDA, ABC resp.
The problem is equivalent to showing that <I1 I2 I3 = 90 deg.
(rest of the proof is obvious)

Now, <I2 D I1 = <I2 D C - <I1 D C = a/2
Also, <I2 C I1 = <I1 C D - <I2 C D = a/2
Thus, the quadrilateral D I2 I1 C is cyclic.

This implies that <D I1 I2 = <D C I2 = b/2
Similarly, we get <B I1 I3 = a/2

Thus, <I2 I1 I3 = <D I1 B - <D I1 I2 - <B I1 I3
= 90+(a+b)/2 - a/2 - b/2 = 90 deg.
as <D I1 B = 90 + (a+b)/2 from triangle BCD.

This completes the proof.

Well, I guess this proof works for all cases (concerning sizes of
a,b,c,d). This can be checked very easily (as in most geometry
problems).

Bhaskara Aditya.

[Estoroth]

Thanks :)