Hình
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Xét ΔHBA và ΔABC có:
\(\widehat{AHB}=\widehat{BAC}=90^o\left(gt\right);\widehat{B}:chung\)
=> ΔHBA ~ ΔABC (g.g)
=> \(\dfrac{HB}{AB}=\dfrac{AH}{AC}\Rightarrow\dfrac{AH}{HB}=\dfrac{AC}{AB}\Rightarrow AH=\dfrac{HB\cdot AC}{AB}\) (*)
Xét ΔHAC và ΔABC có:
\(\widehat{AHC}=\widehat{BAC}=90^o\left(gt\right);\widehat{C}:chung\)
=> ΔHAC ~ ΔABC (g.g)
=> \(\dfrac{HC}{AC}=\dfrac{AH}{AB}\Rightarrow\dfrac{HC}{AH}=\dfrac{AC}{AB}\Rightarrow AH=\dfrac{HC\cdot AB}{AC}\)(**)
Từ (*); (**)
=> \(\dfrac{HB\cdot AC}{AB}=\dfrac{HC\cdot AB}{AC}\)\(\Rightarrow HB\cdot AC^2=AB^2\cdot HC\Rightarrow\left(\dfrac{AC}{AB}\right)^2=\dfrac{HC}{HB}=\dfrac{9}{4}\)
\(\Rightarrow\dfrac{AC}{AB}=\dfrac{3}{2}\)
Vì AD là p/g góc A nên:
\(\dfrac{DC}{DB}=\dfrac{AC}{AB}=\dfrac{3}{2}\)
Vậy.................................................
