已知:如图.AB=AC,AE=AD,BD=EC,求证:∠BAE=∠DAC

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/05 00:10:24
已知:如图.AB=AC,AE=AD,BD=EC,求证:∠BAE=∠DAC
xN0W*!1NI v'!a@3iNH&q`ƥ8mԇ)8)&1r_/,닟{aej4Y7Nw)ֽqx7R /w ܍$J3t*|"I1 T4*CR,35P (8p<{4IUcJNe"|S'}@aD)(gi6wis%&k]EhLbj`~ycvr'tvp_,)eR΃2?=sI&Y˪Ms9ڿS1~ЅU4W1v̰\;R

已知:如图.AB=AC,AE=AD,BD=EC,求证:∠BAE=∠DAC
已知:如图.AB=AC,AE=AD,BD=EC,求证:∠BAE=∠DAC

已知:如图.AB=AC,AE=AD,BD=EC,求证:∠BAE=∠DAC
∵BD=EC
即BE+ED=ED+DC
∴BE=DC
∵AC=AB、AE=AD
∴△AEB≌△ADC(SSS)
∴∠BAE=∠CAD
即∠BAE=∠DAC

BD=EC BE+ED=ED+DC
∴BE=DC
又∵AC=AB、AE=AD
∴△AEB≌△ADC(SSS)
∴∠BAE=∠CAS ∠ BAE=∠DAC 得证

∵BD=EC所以BD-ED=EC-ED∴BE=DC
又∵AB=AC,AE=AD
∴△ABE≌△ACD(SSS)
∴角BAE=角DAC