Problem:
Triangle ABC is inscribed in circle ω with AB=5,BC=7, and AC=3. The bisector of angle A meets side BC at D and circle ω at a second point E. Let γ be the circle with diameter DE. Circles ω and γ meet at E and a second point F. Then AF2=nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let M be the midpoint of BC. Because AE bisects ∠BAC, point E is the midpoint of BC and thus line ME is the perpendicular bisector of BC. Then line ME passes through the center W of circle ω. Therefore ∠EMD=90∘ and M lies on γ. Let rays AM and FM intersect ω at F1 and A1, respectively. Because EFDM is cyclic,
2∠A1WE=∠A1FE=∠MFE=∠MDE=∠BDE=2∠BWE+∠AWC
implying that BA1=AC. Therefore A1 and A are symmetric across line ME, from which it follows that F1 and F are also symmetric across line ME and BF1=CF. Thus ∠BAF1=∠FAC and
∠ABF=∠ABC+∠CBF=2∠AWC+∠CWF=2∠AWC+∠BWF1=∠AMC.
Let R denote the radius of ω. By the Extended Law of Sines, AF=2Rsin∠ABF=2Rsin∠AMC. Applying the Law of Sines to △AMC gives
sin∠AMCAC=sin∠ACBAM or sin∠AMC=AMACsin∠ACB.
Consequently,
AF=AM2R⋅ACsin∠ACB=AMAB⋅AC
by the Extended Law of Sines. By the formula for the length of a median,
