a. Xét △HBA và △ABC có :
\(\hat{B}\) chung
\(\hat{AHB}=\hat{BAC}=90^o\)
\(\Rightarrow\Delta HBA\text{ഗ}\Delta ABC\left(g.g\right)\)
b. Từ a. \(\Rightarrow\dfrac{BH}{AB}=\dfrac{AB}{BC}\left(a\right)\)
\(\Rightarrow AB^2=BH.BC\left(đpcm\right)\)
+) Ta có : \(BC^2=AB^2+AC^2\left(Pytago\right)\)
\(\Leftrightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=5\left(cm\right)\)
+) Mặt khác : \(\left(a\right)\Leftrightarrow BH=\dfrac{AB^2}{BC}=\dfrac{3^2}{5}=1,8\left(cm\right)\)
+) Mà : \(BC=HB+HC\)
\(\Leftrightarrow HC=BC-HB=5-1,8=3,2\left(cm\right)\)
Vậy : \(BH=1,8cm;HC=3,2cm\)
c. Ta có : \(AB^2=AH^2+HB^2\)
\(\Leftrightarrow AH=\sqrt{AB^2-HB^2}=\sqrt{3^2-1,8^2}=2,4\left(cm\right)\)
Xét △AMK và △ABH có :
\(\hat{AKM}=\hat{AHB}=90^o\)
\(\hat{AMK}=\hat{ABH}\) (đồng vị)
\(\Rightarrow\Delta AMK\text{ഗ}\Delta ABH\left(g.g\right)\)
\(\Rightarrow\dfrac{AM}{MB}=\dfrac{AK}{AH}\)
\(\Rightarrow\dfrac{AM}{3}=\dfrac{1,2}{2,4}\Leftrightarrow AM=\dfrac{1,2.3}{2,4}=1,5\left(cm\right)\)
Do MN // BC (gt) \(\Rightarrow\dfrac{AM}{AB}=\dfrac{MN}{BC}\) (hệ quả của định lí Ta-lét)
\(\Leftrightarrow MN=\dfrac{AM.BC}{AB}=\dfrac{1,5.5}{3}=2,5\left(cm\right)\)
Ta lại có : \(AH=AK+HK\)
\(\Leftrightarrow HK=AH-AK=2,4-1,2=1,2\left(cm\right)\)
\(\Rightarrow S_{BMNC}=\dfrac{1}{2}\cdot\left(MN+BC\right)\cdot HK=\dfrac{1}{2}\cdot\left(2,5+5\right)\cdot1,2=4,5\left(cm^2\right)\)
Vậy : ...
