20/09/16 18:27:04.63 s4jUziKT.net
∠CBD = β,
BC = 2R,
とおく。題意より
β + γ = ∠BEC = ε,
⊿ABC = AB･AC/2 = RR sin(2γ),
⊿BCD = BD･CD/2 = RR sin(2β),
ΔABE = AB^2 /(2tan(ε)) = RR{1-cos(2γ)}/tan(ε),
ΔCDE = CD^2 /(2tan(ε)) = RR{1-cos(2β)}/tan(ε),
この4つを足して2で割ると
S = RR{cos(ε) - cos(2ε)cos(β-γ)}/sin(ε),