Xét \(\Delta ABC\)\(\perp\) tại \(A\)
Áp dụng định lí py - ta - go :
BC2 = AB2 + AC2
BC2 = 32 + 42
BC2 = 9 + 16
BC2 = 25
BC = 5 cm
Vậy BC = 5 cm .
Xét \(\Delta ABC\)có BD là đường phân giác \(\widehat{B}\)
\(\Rightarrow\)\(\frac{DA}{DC}=\frac{AB}{BC}\)\(\Rightarrow\) \(\frac{DA}{DC}=\frac{3}{5}\)\(\Rightarrow\) \(\frac{DA}{3}=\frac{DC}{5}\)\(=\frac{DA+DC}{3+5}=\frac{4}{8}=\frac{1}{2}\)
\(\Rightarrow\)\(\frac{DA}{3}=\frac{1}{2}\)\(\Rightarrow\)\(DA=\frac{3}{2}=1,5\)cm
Ta có : AC = AD + DC
4 = 1,5 + DC
\(\Rightarrow DC=2,5\)cm
Xét \(\Delta AHB\) và \(\Delta CAB\) có :
\(\widehat{AHB}\)\(=\)\(\widehat{CAB}\) ( cùng bằng 900 )
\(\widehat{B}\) chung
\(\Rightarrow\)\(\Delta AHB\)\(~\)\(\Delta CAB\) ( g - g )
Do \(\Delta AHB\) \(~\)\(\Delta CAB\)
\(\Rightarrow\)\(\frac{AB}{BH}=\frac{BC}{AB}\)\(\Rightarrow\)\(AB.AB=BH.BC\)\(\Rightarrow\)\(AB^2=BH.BC\)
Ngọc Nguyễn , bạn ơi còn câu c) nữa . trả lời giúp mình câu c) luôn đi
Ta có :
\(S_{\Delta AB\text{D}}=\frac{1}{2}.AB.A\text{D}\)
\(S_{\Delta BC\text{D}}=\frac{1}{2}.AB.C\text{D}\)
\(\Rightarrow\)\(\frac{S_{\Delta AB\text{D}}}{S_{\Delta BC\text{D}}}=\frac{\frac{1}{2}.AB.A\text{D}}{\frac{1}{2}.AB.C\text{D}}=\frac{A\text{D}}{C\text{D}}\)\(\Rightarrow\)\(\frac{S_{\Delta AB\text{D}}}{S_{\Delta BC\text{D}}}=\frac{A\text{D}}{C\text{D}}=\frac{1.5}{2.5}=\frac{3}{5}\)