Problem:
In β³ABC,AB=3,BC=4,CA=5. Circle Ο intersects AB at E and B, BC at B and D, and AC at F and G. Given that EF=DF and EGDGβ=43β, length DE=cabββ, 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 Ο and β DBE=90β,DE is a diameter of Ο. It follows that β DFE=β DGE=90β.
Because DF=EF,β³DFE is an isosceles right triangle and arc DF=arcEF, implying that β DBF=β EBF=45β; that is, BF bisects β ABC. In particular, by the Angle Bisector Theorem, AF<FC.
Because EGGDβ=BCABβ and β DGE=β ABC=90β, conclude that β³DGE and β³ABC are similar. In particular, β GED=β BCA and β GDE=β BAC. Note also that β GED=β GBD and β GDE=β GBE because BDGE is cyclic. Hence β GBC=β GBD=β GED=β BCA=β BCG and β GBA=β GBE=β GDE=β BAC=β BAG, from which it follows that both β³ABG and β³CBG 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, β DGC=β DEF=45β. The Extended Law of Sines and addition formulas imply
