Hình tự vẽ.
a, \(Xét\text{ }\)\(\text{△}ADC\) \(và\) \(\text{△}AEB\) \(có:\)
\(AD=AE\left(gt\right)\)
\(\widehat{A}\) \(Chung\) ⇒ \(\text{△}ADC\) = \(\text{△}AEB\)
\(AB=AC\left(gt\right)\) (c.g.c)
⇒ BE=CD(2 cạnh t.ứ)
b, Có \(\text{△}ADC\) = \(\text{△}AEB\)
⇒ \(\widehat{ACD}=\widehat{ABE}\)
Có AB=AC
Mà AD=AE
⇒ AB-AD=AC-AE
⇒ BD=CE
Xét △\(KBD\) và △\(KCE\) có
\(\widehat{KBD}=\widehat{KCE}\) (cmt)
BD=CE(cmt)
\(\widehat{KDB}=\widehat{KEC}\) ( Do \(\widehat{KBD}=\widehat{KCE}\) , \(\widehat{DKB}=\widehat{EKC}\)(đđ) )
⇒ △\(KBD\) = △\(KCE\)
c,
