1,在△ABC中,D,E分别是∠ACB与∠ABC的三等分线的交点,若∠ABC+∠ACB=120°,求∠CDE
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 19:25:37
![1,在△ABC中,D,E分别是∠ACB与∠ABC的三等分线的交点,若∠ABC+∠ACB=120°,求∠CDE](/uploads/image/z/13893402-66-2.jpg?t=1%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CD%2CE%E5%88%86%E5%88%AB%E6%98%AF%E2%88%A0ACB%E4%B8%8E%E2%88%A0ABC%E7%9A%84%E4%B8%89%E7%AD%89%E5%88%86%E7%BA%BF%E7%9A%84%E4%BA%A4%E7%82%B9%2C%E8%8B%A5%E2%88%A0ABC%2B%E2%88%A0ACB%3D120%C2%B0%2C%E6%B1%82%E2%88%A0CDE)
1,在△ABC中,D,E分别是∠ACB与∠ABC的三等分线的交点,若∠ABC+∠ACB=120°,求∠CDE
1,在△ABC中,D,E分别是∠ACB与∠ABC的三等分线的交点,若∠ABC+∠ACB=120°,求∠CDE
1,在△ABC中,D,E分别是∠ACB与∠ABC的三等分线的交点,若∠ABC+∠ACB=120°,求∠CDE
∵在△BCD中,EC平分∠DCB,EB平分∠CBD
∴DE平分∠BDC
∵∠ A=60°,
∴∠ABC+∠ACB=180-60=120°
∵点D、E分别是<ACB与<ABC三等分线的交点
∴∠BCD=2/3∠ACB,∠CBD=2/3∠ABC
∴∠BCD+∠CBD=2/3*120°=80°
∴∠BCD=180-(∠BCD+∠CBD)=100°
∴∠CDE=50°
∵∠ C=∠ EC'F,∠C'EF=∠CEF,∠C'FE=∠CFE,
∠C'EF+∠EC'F+∠C'FE=180°,∠C'EF+∠1+∠C'FE+∠2+∠CEF+∠CFE=360°,
∴2∠C'EF+∠1+2∠C'FE+∠2=360°,2∠C'EF+2∠C'FE+2∠C=360°,
∴∠1+∠2=2∠ C