Question

如图，在三角形ABC中，D是BC边上的一点，若角BAD=角C=2角DAC=45°，DC=2,求BD的长

不用正弦定理回答

Answer 1

∵∠BAD=∠C=2∠DAC=45°
∴∠DAC==45°/2
∴∠BAC=∠BAD+∠DAC=45°+45°/2=67.5°
∴∠B=180°-∠BAC-∠C=180°-45-67.5=67.5°
由正弦定理得
DC/sin22.5°=AD/sin45°
AD=DCsin45°/sin22.5°
由正弦定理得
BD/sin45°=AD/sin67.5°
BD=ADsin45°/sin67.5°
=DCsin²45°/sin22.5°sin67.5°
=DCsin²45°/sin22.5°cos22.5°
=2DCsin²45°/2sin22.5°cos22.5°
=2DCsin²45°/sin45°
=2DCsin45°
=2×2×√2/2
=2√2

Answer 2

角BAD=角C,三角形BAD和三角形BCA相似，AB:BC:AC=BD:AB:AD
若角BAD=角C=2角DAC=45°,角B=180°-5*45°/2=135°/2
角ADC=180°-3*45°/2=225°/2,角ADB=180°-225°/2=135°/2=角B,所以AB=AD，则AC=BC
AB:BC=BD:AB=AB:(BD+DC),AB^2=BD(BD+2),
BD^2=2AB^2-2AB^2cos45°=2AB^2(1-√2/2)
BD^2/(2-√2)=BD^2+2BD,(1-√2)BD^2+2(2-√2)BD=0,BD=2(2-√2)/(√2-1)=2(3√2-4)