Gọi $E$ là giao của tia phân giác $\widehat{C}, \widehat{D}$

$\Rightarrow$ $\widehat{ECD}=\widehat{ECB}=\dfrac{1}{2}\widehat{DCB}=65^o; \widehat{EDC}=\widehat{EDA}=\dfrac{1}{2}\widehat{ADC}=55^o$

Xét $\triangle$ $EBC$ và $\triangle$ $EDC$ có:

+ $BC=DC$ (gt)

+ $\widehat{ECB}=\widehat{ECD}$ (cmt)

+ $EC$ chung

$\Rightarrow$ $\triangle$ $EBC$ $=$ $\triangle$ $EDC$ (C-G-C)

$\Rightarrow$ $\widehat{EBC}=\widehat{CDE}=55^o$

và $\widehat{DEC}=\widehat{BEC}$ 

$\triangle$ $ECD$ có $\widehat{ECD}=65^o; \widehat{EDC}=55^o$

$\Rightarrow$ $\widehat{DEC}=180^o-65^o-55^o=60^o$

$\Rightarrow$ $\widehat{DEC}=\widehat{BEC}=60^o$

CMTT ta được: $\widehat{DEC}=\widehat{DEA}=60^o$

$\Rightarrow$ $\widehat{DEC}+\widehat{BEC}+\widehat{DEA}=180^o$

$\Rightarrow$ $3$ điểm $B, E, A$ thẳng hàng

Mà $\widehat{EBC}=55^o$

$\Rightarrow$ $\widehat{ABC}=55^o$

Vậy $\widehat{ABC}=55^o$