ABC is a triangle. BD is the an angle bisector. E, F are the feet of the perpendiculars from A, C respectively to the line BD. M is the foot of the perpendicular from D to the line BC. Show that ∠DME = ∠DMF.
Solution
Let H be the foot of the perpendicular from D to AB. ∠AHD = ∠AED = 90o, so AHED is cyclic. Hence ∠DAE = ∠DHE. But M is the reflection of H is the line BD, so ∠DME = ∠DAE.
AE is parallel to CD, so ∠DAE = ∠DCF. ∠DFC = ∠DMC, so DMCF is cyclic. Hence ∠DCF = ∠DMF. Hence ∠DME = ∠DMF.
© John Scholes
jscholes@kalva.demon.co.uk
19 Oct 2002