< PREVIOUS PROBLEM | NEXT PROBLEM >

The figure below shows a triangle ABC with a cevian BD. DE is the bisector of angle ADB, CE and BD meet at O, AO extended meets BC at F. Prove that (1) DF is the bisector of angle BDC; (2) angle EDF = 90 degrees.

Home | Search | Geometry | Problems | All Problems | Open Problems | Visual Index | 10 Problems | 971-980 | Triangles | Angle Bisector | Concurrency | Perpendicular lines | Email | Solution / comment | By Antonio Gutierrez