Question

已知：如图，AB=AC=BD,AD=CD,求：角BAC的角度。

Answer 1

因为AB=AC，角B=角C，设角B=角C=角1
因为AD=DC，角C=角DAC=角1
因为AB=BD，角BAD=角BDA，设角BAD=角BDA=角2
则角2=角BDA=角C+角DAC=2角1
则在三角形BAC中，角B+角C+角BAC=180°
即角B+角C+角BAD+角DAC=180°
即角1+角1+角2+角1=180°
即5角1=180°
角1=36°
则角BAC=角BAD+角DAC=角2+角1=3角1=108°

Answer 2

∠ADB=∠DAC+∠DCA AB=BD ∠BAD=∠BDA AB=AC ∠ABC=∠ACB AD=CD ∠DCA=∠DAC
∠BAC+∠ABC+∠ACB=180 5∠ABC=180,∠ABC=36, ∠BAC=3∠ABC=108

Answer 3

设∠B为x，∠C=∠DAC=x，∠BDA=∠BAD=180°-3x，在△ABD中列角度方程，解得x=36°，∠BAC=108°