Problem: In acute triangle ABC the bisector of β A meets side BC at D. The circle with center B and radius BD intersects side AB at M; and the circle with center C and radius CD intersects side AC at N. Then it is always true that
Answer Choices:
A. β CND+β BMDββ DAC=120β
B. AMDN is a trapezoid
C. BC is parallel to MN
D. AMβAN=23(DBβDC)β
E. ABβAC=23(DBβDC)β
Solution:
In the adjoining figure CDBDβ=ACABβ, since the bisector of an angle of a triangle divides the opposite side into segments which are proportional to the two adjacent sides. Since CN=CD and BM=BD, we have CNBMβ=ACABβ, which implies MN is parallel to CB. Statements (A), (B), (D) and (E) are false if
β A=90ββΞΈ,β B=60β and β C =30β+ΞΈ, where ΞΈ is any sufficiently small positive angle.
