a) Xét △ABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow\widehat{BAC}+50^o+30^o=180^o\\ \Rightarrow\widehat{BAC}=100^o\)
b) Xét △AHB và △DHB có:
\(HA=HD\\ \widehat{AHB}=\widehat{DHB}=90^o\\ BH:\text{ cạnh chung}\)
\(\Rightarrow\text{△AHB = △DHB (c.g.c)}\)
\(\Rightarrow\widehat{HAB}=\widehat{HDB}\left(\text{hai góc tương ứng}\right)\) (1)
Xét △AHC và △DHC có:
\(HA=HD\\ \widehat{AHC}=\widehat{DHC}=90^o\\ CH:\text{cạnh chung}\)
\(\Rightarrow\text{△AHC = △DHC (c.g.c)}\)
\(\Rightarrow\widehat{CAH}=\widehat{CDH}\left(\text{hai góc tương ứng}\right)\) (2)
(1) (2) \(\Rightarrow\widehat{BAC}=\widehat{BDC}\)
