Let D and E be points on sides AB and AC respectively of ΔABC.
Let the bisectors of /ABE and /ACD intersect in point F.

Prove that /BDC + /BEC = 2 /BFC.

Let angle ABE = 2x and angle ACD = 2y. Let G, H and I be the points of intersection of BF and CD, CF and BE, BE and CD respectively.

Using D and E to denote the angles BDC and BEC, it follows, by using external angles of triangles etc. that:

angle GIH = 2x + D = 2y + E      therefore angle GIH = x + y + (D + E)/2

angle FGI = 180⁰ - x - D

angle FHI = 180⁰ - y - E

In quadrilateral FGIH:

angle BFC    = 360⁰ - angle FGI - angle FHI - angle GIH

            = 360⁰ - (180⁰ - x - D) - (180⁰ - y - E) - (x + y + D/2 + E/2)

            = D/2 + E/2

Therefore          D + E = 2(angle BFC)   as required

Posted by Harry on 2010-08-03 20:16:17