Hình tự vẽ nhé
Ta có: \(\widehat{C}=30^o\Rightarrow\widehat{B}=60^o\)(phụ nhau)
\(\widehat{B}+\widehat{BAD}+\widehat{BDA}=180^o\)(tổng 3 góc tam giác)
\(60^o+\widehat{BAD}+\widehat{BDA}=180^o\)
\(\widehat{BAD}+\widehat{BDA}=180^o-60^o=120^o_{\left(1\right)}\)
Mà BD=BA => \(\widehat{BAD}=\widehat{BDA}_{\left(2\right)}\)
Từ (1),(2) => \(\widehat{BAD}=\widehat{BDA}=\dfrac{120^o}{2}=60^o\)
\(\widehat{B}=\widehat{BAD}=\widehat{BDA}=60^o\)
=> \(\Delta ABD\) đều
\(\widehat{BAD}+\widehat{DAC}=\widehat{A}=90^o\)
\(60^o+\widehat{DAC}=90^o\)
\(\widehat{DAC}=90^o-60^o=30^o\)
b)Ta có: \(\widehat{DAC}=\widehat{C}=30^o\)
\(\Rightarrow\Delta ADC\) cân tại D
\(\Rightarrow DA=DC\)
Xét \(\Delta ADE\) và \(\Delta CDE:\)
\(\widehat{AED}=\widehat{DEC}=90^o\)
\(DA=DC\left(cmt\right)\)
\(\widehat{DAC}=\widehat{C}=30^o\)
\(\Rightarrow\Delta ADE=\Delta CDE\)(ch-gn)
c)
\(BD=AD=5cm\left(cma\>\right)\)
\(DC=AD\left(cmb\right)\)
\(\Rightarrow BD=DC\)
Mà \(BD+DC=BC\)
\(\Rightarrow BC=2BD=2.5=10cm\)
Xét \(\Delta ABC\)
\(BC^2=AB^2+AC^2\)
\(10^2=5^2+AC^2\)
\(\Rightarrow AC^2=10^2-5^2=100-25=75\)
\(\Rightarrow AC=\sqrt{75}cm\)
d) Bạn xem lại câu này chứ mình thấy nó sai á
