Đáp án+Giải thích các bước giải:
$a)$Xét $\Delta HBA$ và $\Delta ABC$
$\widehat{B}$: chung
$\widehat{AHB}=\widehat{CAB}\\ \Rightarrow \Delta HBA \backsim \Delta ABC$
$b)\Delta ABC$ vuông tại $A$
$\Rightarrow BC=\sqrt{AB^2+AC^2}=20(cm)$
$\Delta ABC$ vuông tại $A$, đường cao $AH$
$\Rightarrow AB.AC=AH.BC\\ \Rightarrow AH=\dfrac{AB.AC}{BC}=9,6(cm)$
$\Delta AHB$ vuông tại $H$
$\Rightarrow BH=\sqrt{AB^2-AH^2}=7,2(cm)$
$c)AD$ là phân giác $\widehat{BAC}$
$\Rightarrow \dfrac{BD}{CD}=\dfrac{AB}{AC}=\dfrac{3}{4}\\ \Leftrightarrow \dfrac{BD}{CD}=\dfrac{3}{4}\\ \Leftrightarrow \dfrac{BD}{BC-BD}=\dfrac{3}{4}\\ \Leftrightarrow 4BD=3(20-BD)\\ \Leftrightarrow 4BD=60-3BD\\ \Leftrightarrow 7BD=60\\ \Leftrightarrow BD=\dfrac{60}{7}(cm)\\ \Rightarrow CD=BC-BD=\dfrac{80}{7}(cm)\\ d)S_{ABC}=\dfrac{1}{2}AB.AC=96(cm^2)\\ \Delta AHC, KN//HC\\ \Rightarrow \dfrac{AN}{AC}=\dfrac{AK}{AH}=\dfrac{3}{8}\\ MN//BC\\ \Rightarrow \widehat{M_1}=\widehat{B}; \widehat{N_1}=\widehat{C}$
Xét $\Delta AMN$ và $\Delta ABC$
$\widehat{M_1}=\widehat{B}\\ \widehat{N_1}=\widehat{C}\\ \Rightarrow \Delta AMN \backsim \Delta ABC\\ \Rightarrow \dfrac{S_{AMN}}{S_{ABC}}=\left(\dfrac{AN}{AC}\right)^2=\dfrac{9}{64}\\ \Rightarrow S_{AMN}=\dfrac{9}{64}S_{ABC}=\dfrac{27}{2}(cm^2)\\ S_{BMNC}=S_{ABC}-S_{AMN}=\dfrac{165}{2}(cm^2)$
