Hình:
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a/ Áp dụng đl pitago ta có:
\(BC^2=AB^2+AC^2=6^2+8^2=100\Rightarrow BC=10\left(cm\right)\)
Vì BD là p/g góc ABC nên:
\(\dfrac{AD}{DC}=\dfrac{AB}{BC}\Rightarrow\dfrac{AD}{AB}=\dfrac{DC}{BC}=\dfrac{AD+DC}{AB+BC}=\dfrac{AC}{6+10}=\dfrac{8}{16}=\dfrac{1}{2}\)
=> \(\left\{{}\begin{matrix}AD=\dfrac{1}{2}\cdot6=3\left(cm\right)\\DC=\dfrac{1}{2}\cdot10=5\left(cm\right)\end{matrix}\right.\)
b/ Xét ΔABC và ΔHBA có:
\(\widehat{BAC}=\widehat{AHB}=90^o\)
\(\widehat{ABC}:chung\)
=> Xét ΔABC ~ ΔHBA(g.g)
=> \(\dfrac{AB}{HB}=\dfrac{BC}{AB}\Rightarrow AB^2=HB\cdot BC\left(đpcm\right)\)
c/ Vì: ΔABC ~ ΔHBA => \(\widehat{BAH}=\widehat{C}\)
Xét ΔABI và ΔCBD có:
\(\widehat{BAH}=\widehat{C}\left(cmt\right);\widehat{ABI}=\widehat{CBD}\left(gt\right)\)
=> ΔABI ~ ΔCBD (g.g) (đpcm)
d/ Vì I ∈ BD => BI là p/g góc ABC
=> \(\dfrac{IH}{IA}=\dfrac{HB}{AB}\) (1)
Ta có: ΔHBA ~ ΔABC
=> \(\dfrac{HB}{AB}=\dfrac{AB}{BC}\) (2)
BD là p/g góc ABC
=> \(\dfrac{AD}{DC}=\dfrac{AB}{BC}\) (3)
Từ (1), (2), (3) => \(\dfrac{IH}{IA}=\dfrac{AD}{DC}\) (đpcm)
