Problem:
Triangle ABC has side lengths AB=4,BC=5, and CA=6. Points D and E are on ray AB with AB<AD<AE. The point F=C is a point of intersection of the circumcircles of △ACD and △EBC satisfying DF=2 and EF=7. Then BE can be expressed as da+bc, where a,b,c, and d are positive integers such that a and d are relatively prime, and c is not divisible by the square of any prime. Find a+b+c+d.
Solution:
Because quadrilateral DFAC is cyclic, ∠DFC=∠DAC=∠BAC. Because quadrilateral EFBC is cyclic, ∠EFC=∠EBC=180∘−∠ABC.
Hence ∠EFD=∠EFC−∠DFC=180∘−∠ABC−∠BAC=∠ACB. Applying the Law of Cosines to △DEF and △ABC gives
cos∠EFD=cos∠ACB=2⋅6⋅562+52−42=43
and
DE=72+22−2⋅2⋅7⋅43=42
Let CF intersect line AB at G. Because △ACG∼△FDG and △BCG∼△FEG,
3=FDAC=DGCG=GFGA and 75=FEBC=EGCG=GFGB.
Therefore
521=DGEG=DGED+DG and 521=GBGA=GBGB+BA.
Solving for DG and GB yields GD=452 and BG=45, so
