Problem:
In △ABC,AB=3,BC=4,CA=5. Circle ω intersects AB at E and B, BC at B and D, and AC at F and G. Given that EF=DF and EGDG=43, length DE=cab, where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. Find a+b+c.
Solution:
Because points B,D,E,F,G all lie on ω and ∠DBE=90∘,DE is a diameter of ω. It follows that ∠DFE=∠DGE=90∘.
Because DF=EF,△DFE is an isosceles right triangle and arc DF=arcEF, implying that ∠DBF=∠EBF=45∘; that is, BF bisects ∠ABC. In particular, by the Angle Bisector Theorem, AF<FC.
Because EGGD=BCAB and ∠DGE=∠ABC=90∘, conclude that △DGE and △ABC are similar. In particular, ∠GED=∠BCA and ∠GDE=∠BAC. Note also that ∠GED=∠GBD and ∠GDE=∠GBE because BDGE is cyclic. Hence ∠GBC=∠GBD=∠GED=∠BCA=∠BCG and ∠GBA=∠GBE=∠GDE=∠BAC=∠BAG, from which it follows that both △ABG and △CBG are isosceles with AG=BG=CG, implying that G is the midpoint of side AC.
Note that AF<FC and AG=GC. Hence BDGFE is a convex cyclic pentagon. Because DEFG is cyclic, ∠DGC=∠DEF=45∘. The Extended Law of Sines and addition formulas imply
