Problem:
Points A,B,C,D, and E are equally spaced on a minor arc of a circle. Points E,F,G,H,I, and A are equally spaced on a minor arc of a second circle with center C as shown in the figure below. The angle β ABD exceeds β AHG by 12β. Find the degree measure of β BAG.
Solution:
Because A,I,H,G,F, and E are equally spaced, let Ξ±=β ECF=β FCG=β GCH=β HCI=β ICA. It follows that β ACE=β ABE=β ADE=5Ξ±. Also, AHG=3Ξ± so β AHG=2360ββ3Ξ±β. Because β ACE=5Ξ±,ACE=360ββ10Ξ±, and ABD=270ββ215Ξ±β. Thus β ABD=2360ββ(270ββ215Ξ±β)β=45β+415Ξ±β. Then β ABDββ AHG=(45β+415Ξ±β)β(180ββ23Ξ±β)=421Ξ±ββ135β=12β. Hence Ξ±=28β.
Now EFG=2Ξ±, so β EAG=Ξ±=28β. From above ABD=270ββ215Ξ±β=60β=BCE, so β BAE=260ββ=30β. Finally β BAG=β BAE+β EAG=30β+28β=58ββ.
