b)

Áp dụng định lý Py-ta-go cho tam giác vuông ABC. Ta có:

\(AB^2+AC^2=BC^2\)

\(\Leftrightarrow15^2+20^2=BC^2\)

\(\Leftrightarrow BC=25\)

Ta có: \(\text{ΔABC ∼ ΔHBA }\)   (cm câu a)

\(\Rightarrow\dfrac{AC}{AH}=\dfrac{BC}{AB}=\dfrac{AB}{BH}\)

⇔ \(\dfrac{AH}{AC}=\dfrac{AB}{BC}=\dfrac{BH}{AB}\)

⇔ \(\dfrac{AH}{20}=\dfrac{15}{25}=\dfrac{BH}{15}\)

\(\Rightarrow\left\{{}\begin{matrix}AH=12\\BH=9\end{matrix}\right.\)

⇒ \(CH=BC-BH=25-9=16\)

a)

Xét \(\Delta ABC\) và \(\Delta HBA\) có:

           \(\widehat{B}:chung\)

      \(\widehat{BAC}=\widehat{BHA}\left(=90^o\right)\)

\(\Rightarrow\Delta ABC\sim\Delta HBA\left(g.g\right)\)           \(\left(ĐPCM\right)\)

\(x^2+x^3=0\)

\(\Leftrightarrow x^2\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\end{matrix}\right.\)       \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy...

Ta có: \(\widehat{xOy}=60^o\)

⇔ \(\widehat{xOz}+\widehat{zOy}=60^o\)

Mà \(\widehat{xOy}=3.\widehat{zOy}\)

⇒ \(4.\widehat{zOy}=60^o\)

⇔ \(\widehat{zOy}=15^o\)

⇒ \(\widehat{xOz}=3.15^o=45^o\)

Lại có: \(\widehat{xOt}=10.\widehat{zOy}\)

⇔  \(\widehat{xOt}=150^o\)

Ta có: \(\widehat{tOz}=\widehat{xOt}-\widehat{xOz}\)

⇔ \(\widehat{tOz}=150^o-45^o\)

⇔ \(\widehat{tOz}=105^o\)

Ơ, s tui k thấy j cả. Câu hỏi đâu?? 

b)

Ta có: \(\Delta ADB\sim\Delta AEC\)   (cm câu a)

⇒ \(\dfrac{AD}{AE}=\dfrac{AB}{AC}\)      

⇔ \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)

Xét \(\Delta ADE\) và \(\Delta ABC\) có:

          \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)

           \(\widehat{A}\) : chung

⇒  \(\Delta ADE\sim\Delta ABC\)      \(\left(c.g.c\right)\)

⇒      \(\widehat{AED}=\widehat{ACB}\)       (2 góc tương ứng)

⇒        \(ĐPCM\)