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 ∠ABD exceeds ∠AHG by 12∘. Find the degree measure of ∠BAG.
Solution:
Because A,I,H,G,F, and E are equally spaced, let α=∠ECF=∠FCG=∠GCH=∠HCI=∠ICA. It follows that ∠ACE=∠ABE=∠ADE=5α. Also, AHG=3α so ∠AHG=2360∘−3α​. Because ∠ACE=5α,ACE=360∘−10α, and ABD=270∘−215α​. Thus ∠ABD=2360∘−(270∘−215α​)​=45∘+415α​. Then ∠ABD−∠AHG=(45∘+415α​)−(180∘−23α​)=421α​−135∘=12∘. Hence α=28∘.
Now EFG=2α, so ∠EAG=α=28∘. From above ABD=270∘−215α​=60∘=BCE, so ∠BAE=260∘​=30∘. Finally ∠BAG=∠BAE+∠EAG=30∘+28∘=58∘​.
