I is the incenter of the triangle ABC. The point D outside the triangle is such DA is parallel to BC and DB = AC, but ABCD is not a parallelogram. The angle bisector of BDC meets the line through I perpendicular to BC at X. The circumcircle of CDX meets the line BC again at Y. Show that DXY is isosceles.
Solution
Let IX meet BC at Z. Then using equal tangents, (BC - CZ) + (AC - CZ) = AB, so CZ = (AC + BC - AB)/2. Suppose the excircle opposite D of DBC touches BC at Z'. Then, again considering equal tangents, DB + (BC - CZ') = DC + CZ', so CZ' = (BD + BC - DC)/2 = (AC + BC - AB)/2 = CZ, so Z' and Z coincide. Since X lies on the perpendicular to BC at Z and on the bisector of ∠BDC, it must also be the center of the excircle. Hence XC is the exterior bisector of ∠BCD. So ∠XCB = 90 - ∠BCD/2.
By construction, YDCX is cyclic, so ∠YDX = ∠YCX = ∠XCB. Also ∠BCD = ∠YCD = ∠YXD. Hence ∠YDX = 90 - ∠YXD/2. Hence YX = DX.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002