Question

如图在三角形ABC中,D是BC上一点,E是AC上一点,且满足AD=AB,角ADE=角C

Answer 1

(1)①∠AED=∠EDC+∠C,∠ADC=∠ADE+∠EDC

∵∠C=∠ADE

∴∠AED=∠ADC

②∵∠DEC=∠DAE+∠ADE且∠C=∠ADE

∴∠DAC+∠C=∠DEC

又∵∠ADB=∠DAC+∠C

∴∠ADB=∠DEC

连接bc

同理可证：△AEB∽△ABC然后AB：AE=AC：AB即AB平方=AE×AC