Question

Giúp mình với ạ :33

Answer 1

a) Xét tam giác ABC và tam giác HBA có:

\(\widehat{BAC}=\widehat{BHA}=90^0\)

\(\widehat{ABC}\) chung

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

\(\Rightarrow\dfrac{AB}{BH}=\dfrac{BC}{AB}\Rightarrow AB^2=BC.BH\)

b) Xét tam giác ABC và tam giác HAC có:

\(\widehat{BAC}=\widehat{AHC}=90^0\)

\(\widehat{ACB}\) chung

\(\Rightarrow\Delta ABC\sim\Delta HAC\left(g.g\right)\)

\(\Rightarrow\dfrac{AC}{HC}=\dfrac{BC}{AC}\Rightarrow AC^2=BC.HC\)

c) Ta có: \(\left\{{}\begin{matrix}\Delta ABC\sim\Delta HBA\left(cmt\right)\\\Delta ABC\sim\Delta HAC\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta HBA\sim\Delta HAC\)

\(\Rightarrow\dfrac{AH}{CH}=\dfrac{BH}{AH}\Rightarrow AH^2=CH.BH\)

d) Ta có: \(\Delta ABC\sim\Delta HAC\left(cmt\right)\)

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