# Geometry Help - Free Math Help

Let K (on an extension of AB) be the center of the circumcircle of triangle OEA, and let the circumradius be r. Let alpha angle CAO angle KAO be the external half-angle at A.
Note that these two parameters are dependent: triangle KAE is isosceles with legs r, base 2, and base angles 2alpha; thus, rcos 2alpha 1. Since triangle AKO is isosceles with angle KOA angle KAO alpha angle CAO, we have OK parallel CA.
This guarantees triangle OKB sim triangle EAB, and we see fracOK2 fracbkab implies fracr2 fracABrAB implies AB frac2 rr-2 (Observe that this requires r 2.) By the Angle Bisector Theorem in triangle ABC, fracaeab fraccecb implies frac2ABfrac3CB implies CB frac3rr-2, via the Law of Cosines at (internal) angle A in triangle ABC, cosleft(pi-2alpharight) fracAC2AB2-BC22acab fracr2-5r5r(r-2 recalling that rcos 2alpha1 (and, of course, that cos(pi-2alpha)-cos 2alpha we have -frac1r fracr2-5r5r(r-2) implies (r-1 r-3)
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