Câu 16:
a. -△ABM và △ACB có: \(\widehat{ABM}=\widehat{ACB};\widehat{A}\) là góc chung.
\(\Rightarrow\)△ABM∼△ACB (g-g)
b. △ABM∼△ACB \(\Rightarrow\dfrac{S_{ABM}}{S_{ACB}}=\left(\dfrac{AB}{AC}\right)^2=\left(\dfrac{2}{4}\right)^2=\dfrac{1}{4}\)
-△ABK và △ACH có: \(\widehat{ABK}=\widehat{ACH};\widehat{AKB}=\widehat{AHC}=90^0\)
\(\Rightarrow\)△ABK∼△ACH (g-g) \(\Rightarrow\dfrac{S_{ABK}}{S_{ACH}}=\left(\dfrac{AB}{AC}\right)^2=\left(\dfrac{2}{4}\right)^2=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{4}=\dfrac{S_{ABM}}{S_{ACB}}=\dfrac{S_{ABK}}{S_{ACH}}=\dfrac{S_{ABM}-S_{ABK}}{S_{ACB}-S_{ACH}}=\dfrac{S_{AKM}}{S_{AHB}}\)
\(\Rightarrow S_{AHB}=4S_{AKM}\)
