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Trên AD lấy E sao cho $\widehat{BCA}=\widehat{ACE}$
CM: $\Delta ABC = \Delta AEC(g-c-g) \Longrightarrow BC=CE$ (1)
và $\widehat{AEC}=\widehat{ABC}=110^o$
Ta có: $\widehat{AEC}+\widehat{CED}=180^o \Longrightarrow \widehat{CED}=180^o- \widehat{AEC}=180^o-110^o=70^o$
$\Longrightarrow \widehat{CED}=\widehat{D} \Longrightarrow \Delta CED$ cân tại C $\Longrightarrow CE=CD$ (2)
Từ (1) và (2) $\Longrightarrow BC=CD$