BC = BD + DC = 2 + 3 = 5 (cm)
DEA = EAF = AFD = 900
=> AEDF là hcn có AD là tia phân giác
=> AEDF là hình vuông
=> \(\left\{\begin{matrix}\text{AF//ED}\\\text{AE//FD}\\DF=ED\end{matrix}\right.\)
Tam giác ABC có AD là tia phân giác
=> \(\frac{AB}{AC}=\frac{DB}{DC}=\frac{2}{3}\) (định lý)
=> \(\left\{\begin{matrix}AB=\frac{2}{3}AC\\AC=\frac{3}{2}AB\end{matrix}\right.\)
Tam giác ABC vuông tại A có:
AB2 + AC2 = BC2 (định lý Pytago)
\(AB^2+\left(\frac{3}{2}AB\right)^2=5^2\)
\(AB=\frac{10\sqrt{13}}{13}\) (cm)
Theo định lý Talet, ta có:
\(\frac{DF}{AB}=\frac{CD}{BC}=\frac{3}{5}\Rightarrow DF=\frac{3}{5}AB=\frac{3}{5}\times\frac{10\sqrt{13}}{13}=\frac{6\sqrt{13}}{13}\left(cm\right)\)
\(\frac{FC}{AC}=\frac{DF}{AB}=\frac{DF}{\frac{2}{3}AC}=\frac{\frac{3}{2}DF}{AC}\Rightarrow FC=\frac{3}{2}DF\)
\(\frac{BE}{AB}=\frac{ED}{AC}=\frac{ED}{\frac{3}{2}AB}=\frac{\frac{2}{3}ED}{AB}\Rightarrow BE=\frac{2}{3}ED\)
\(S_{DEB}=ED\times EB\times\frac{1}{2}=ED\times\frac{2}{3}ED\times\frac{1}{2}=\frac{1}{3}DE^2=\frac{1}{3}DF^2\left(cm^2\right)\)
\(S_{DFC}=DF\times FC\times\frac{1}{2}=DF\times\frac{3}{2}DF\times\frac{1}{2}=\frac{3}{4}DF^2\left(cm^2\right)\)
\(S_{DEB}+S_{DFC}=\frac{3}{4}DF^2+\frac{1}{3}DF^2=\frac{3}{4}\left(\frac{6\sqrt{13}}{13}\right)^2+\frac{1}{3}\left(\frac{6\sqrt{13}}{13}\right)^2=3\left(cm^2\right)\)
