Có $ΔABC$ cân tại $A$
$⇒\widehat{ABC}=\widehat{ACB}$
Hay $\widehat{ABC}=\widehat{DCB}$
$ΔADB$ cân tại $A$
$⇒\widehat{ADB}=\widehat{ABD}$
Hay $\widehat{BDC}=\widehat{ABD}$
Nên $\widehat{ABC}+\widehat{ABD}=\widehat{DCB}+\widehat{BDC}$
Hay $\widehat{CBD}=\widehat{DCB}+\widehat{BDC}$
Mà $\widehat{CBD}+\widehat{DCB}+\widehat{BDC}=180^o$
$⇒2.\widehat{CBD}=180^o$
$⇒\widehat{CBD}=90^o$
$⇒ΔBCD$ vuông tại $B$