连结AE,设AD的垂直平分线交AD于F
可知∠ADE=∠DAE,DE=AE,∠BAD=∠DAC
∠ACE=∠ADE+∠DAC (∠ACE为△ACD的外角)
∠ACE=∠DAE+∠BAD=∠BAE (∠ADE=∠DAE,∠BAD=∠DAC)
∠BEA为公共角
∴△ACE∽△BAE
∴BE/AE=AE/EC
AE²=BE*EC
即DE²=BE*EC (DE=AE)
一道数学题,证相似的因为图画不了,我就稍微描述一下.∠B大概是40°左右∠ADB大概120°左右∠BAC大概30°左右题
连结AE,设AD的垂直平分线交AD于F
可知∠ADE=∠DAE,DE=AE,∠BAD=∠DAC
∠ACE=∠ADE+∠DAC (∠ACE为△ACD的外角)
∠ACE=∠DAE+∠BAD=∠BAE (∠ADE=∠DAE,∠BAD=∠DAC)
∠BEA为公共角
∴△ACE∽△BAE
∴BE/AE=AE/EC
AE²=BE*EC
即DE²=BE*EC (DE=AE)