AD=BD,得到∠DBA=∠DAB
AD=CD,得到∠DCA=∠DAC
由三角形内角和为180°,得到∠DBA+∠DCA+∠BAC=180°
∠DBA+∠DCA+∠DAB+∠DAC=180°
2*(∠DAB+∠DAC)=180°
∠DAB+∠DAC=90°,得到∠BAC=90°.
AD=BD,得到∠DBA=∠DAB
AD=CD,得到∠DCA=∠DAC
由三角形内角和为180°,得到∠DBA+∠DCA+∠BAC=180°
∠DBA+∠DCA+∠DAB+∠DAC=180°
2*(∠DAB+∠DAC)=180°
∠DAB+∠DAC=90°,得到∠BAC=90°.