Question

三角形中∠ABC=∠ACB,D为BC上一点,E为直线AC上一点,且∠ADE=∠AED;求证:∠BAD=2∠CDE

急用

Answer 1

∠CDE+∠ACB=∠AED(因为∠ABC=∠ACB)
所以∠CDE+∠ABC=∠AED
∠ABC+∠BAD=∠ADC=∠ADE+∠EDC(因为∠ADE=∠AED)
∠ABC+∠BAD=∠AED+∠EDC=∠EDC+∠ABC+∠EDC
即∠ABC+∠BAD=∠CDE+∠ABC+∠CDE消去∠ABC
等∠BAD=∠CDE+∠CDE=2∠CDE

Answer 2

(1)∵∠ABC=∠ACB
∴∠C=∠B
∵∠ADB+∠ADE+∠CDE=180°
∠CED+∠AED=180°
又∵∠ADE=∠AED
∴∠CED=∠ADB+∠CDE
∵∠ADB+∠B+∠BAD=180°
∠ADB+∠C+∠BAD=180°

∠CDE+∠CED+∠C=180°
∠CDE+∠ADB+∠CDE(∠CED=∠ADB+∠CDE)+∠C=180°

∴∠BAD=2∠CDE(去掉相同的：∠ADB and ∠C)

(2)
不知道什么是反向延长线.但是第一解出来对第二也有帮助吧.....

Answer 3

有图才会有真相- -。