a/ Áp dụng định lý pytago có:
\(BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)
Vì BD là p/g góc ABC (gt) nên:
\(\dfrac{AD}{AB}=\dfrac{DC}{BC}=\dfrac{AD+DC}{AB+BC}=\dfrac{AC}{AB+BC}=\dfrac{8}{6+10}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}AD=\dfrac{1}{2}\cdot AB=\dfrac{1}{2}\cdot6=3\left(cm\right)\\DC=\dfrac{1}{2}\cdot BC=\dfrac{1}{2}\cdot10=5\left(cm\right)\end{matrix}\right.\)
b/ Xét \(\Delta ABC\)và \(\Delta HBA\) có:
\(\widehat{A}=\widehat{H}=90^o\)
\(\widehat{B}\)là góc chung
\(\Rightarrow\Delta ABC\)đồng dạng với \(\Delta HBA\) (gg)
⇒ $\frac{AB}{BH}=\frac{BC}{AB}$
⇒ AB²= BH.BC (đpcm)
c/ Xét ΔABC và ΔHBA có:
\(\widehat{BAC}=\widehat{AHC}=90^o\left(gt\right)\)
\(\widehat{B}:chung\)
=> ΔABC ~ ΔHBA (g.g)
=> \(\widehat{A_1}=\widehat{C}\)
Xét ΔABI và ΔCBD có:
\(\widehat{A_1}=\widehat{C}\left(cmt\right)\)
\(\widehat{B_1}=\widehat{B_2}\left(gt\right)\)
=> ΔABI ~ ΔCBD (g.g)
d/ Vì ΔABI ~ ΔCBD => \(\dfrac{IA}{DC}=\dfrac{IB}{BD}\) (1)
Xét ΔIHB và ΔDAB có:
\(\widehat{B_1}=\widehat{B_2}\left(gt\right)\)
\(\widehat{BAD}=\widehat{BHI}=90^o\left(gt\right)\)
=> ΔIHB ~ ΔDAB (g.g)
=> \(\dfrac{IH}{AD}=\dfrac{IB}{BD}\) (2)
Từ (1),(2) => \(\dfrac{IH}{AD}=\dfrac{IA}{DC}\left(=\dfrac{BI}{BD}\right)\)
hay \(\dfrac{IH}{IA}=\dfrac{AD}{DC}\left(đpcm\right)\)
⇒ IH.DC=IA.AD (đpcm)
