已知△ABC中, BD, CE分別是∠B, ∠C的內(nèi)角平分線, BD = CE, 求證AB = AC.設(shè)∠B = 2β, ∠C = 2γ, 在△EBC中由正弦定理得:BC/CE = sin∠CEB/sin∠B = sin(180°-2β-γ)/sin2β = sin(2β+γ)/sin2β.同理在△DBC得:...額...能不能把它全部講一遍,我才初中,三角函數(shù)的公式都還沒學(xué),謝謝了.如果正弦和角公式都沒學(xué), 其實(shí)不太適合看這個證明.這里僅介紹一下用到的公式, 證明就不說了.二倍角公式: sin2A = 2sinAcosA.積化和差公式(之一): 2sinAcosB = sin(A+B)+sin(A-B).差化積公式: sinA-sinB = 2sin((A-B)/2)cos((A+B)/2),cosA-cosB = 2sin((B-A)/2)sin((A+B)/2).接前文, 去分母得sin2βsin(β+2γ)-sin2γsin(2β+γ) = 0.用二倍角公式展開sin2β與sin2γ即得:2sinβcosβsin(β+2γ)-2sinγcosγsin(2β+γ) = 0.對cosβsin(β+2γ)積化和差得: 2sin(β+2γ)cosβ = sin(2(β+γ))+sin2γ.同理2sin(2β+γ)cosγ = sin(2(β+γ))+sin2β.代回得sinβ(sin(2(β+γ))+sin2γ)-sinγ(sin(2(β+γ))+sin2β) = 0.即sin(2(β+γ))(sinβ-sinγ)+sinβsin2γ-sinγsin2β = 0.再用二倍角: sinβsin2γ-sinγsin2β = 2sinβsinγ(cosγ-cosβ).于是有sin(2(β+γ))(sinβ-sinγ)+2sinβsinγ(cosγ-cosβ) = 0.由差化積, sinβ-sinγ = 2sin((β-γ)/2)cos((β+γ)/2).而cosγ-cosβ = 2sin((β-γ)/2)sin((γ+β)/2).代回得2sin((β-γ)/2)·(sin(2(β+γ))cos((β+γ)/2)+2sinβsinγsin((γ+β)/2)) = 0.由0 < 2(β+γ) < 180°, 0 < β, γ < 90°, 0 < (β+γ)/2 < 90°, 可知sin(2(β+γ))cos((β+γ)/2)+2sinβsinγsin((γ+β)/2) > 0.于是sin((β-γ)/2) = 0, 又-90° < (β-γ)/2 < 90°, 只有(β-γ)/2 = 0, 即β = γ.因此∠B = 2β = 2γ = ∠C, 即有AB = AC.