a/Xét tgiac AHB và BCD có
\(\widehat{AHB}=\widehat{BCD}=90\),\(\widehat{ABH}=\widehat{BDC}\left(SLT\right)\)
Suy ra \(\Delta AHB\sim\Delta BCD\left(g-g\right)\)(1)
b/Từ (1) suy ra \(\frac{AH}{BC}=\frac{HB}{DC}\Leftrightarrow\frac{AH}{HB}=\frac{BC}{DC}\left(2\right)\)
Lại có CE là ph/giác nên \(\frac{BC}{DC}=\frac{EB}{DE}\left(3\right)\)
Từ (2) và (3) suy ra \(\frac{AH}{HB}=\frac{EB}{DE}\Rightarrow AH.DE=HB.EB\)
c/Áp dụng Pitago có: \(BD^2=AB^2+AD^2\Leftrightarrow BD=\sqrt{8^2+6^2}=10cm\)
\(\Delta ADB\sim\Delta HDA\left(g-g\right)\left(2\right)\Rightarrow\frac{AD}{HD}=\frac{BD}{AD}\Leftrightarrow AD^2=HD.BD\Leftrightarrow HD=\frac{6^2}{10}=3,6cm\)
Có CE là ph/giác nên \(\frac{EB}{DE}=\frac{BC}{DC}=\frac{6}{8}=\frac{3}{4}\Leftrightarrow DE=\frac{4}{3}EB\)
Ta có DE+EB=\(\frac{4}{3}EB+EB=\frac{7}{3}EB=BD\Rightarrow EB=\frac{30}{7}cm\)
Vậy HE=\(BD-EB-HD=10-\frac{30}{7}-3,6=\frac{74}{35}cm\)
Cũng từ (2) suy ra \(\frac{AB}{AH}=\frac{BD}{AD}=\frac{10}{6}=\frac{5}{3}\Rightarrow AH=\frac{3}{5}AB=\frac{3}{5}.8=4,8cm\)
Ta có \(S_{AHE}=\frac{1}{2}.4,8.\frac{74}{35}=\frac{888}{175}cm\)
Kẻ CK vuôn góc BD dễ dàng CM tgiac AHD =CKB suy ra AH=CK. \(\frac{S_{AHE}}{S_{HEC}}=\frac{\frac{1}{2}AH.HE}{\frac{1}{2}CK.HE}=\frac{AH}{CK}\Rightarrow S_{AHE}=S_{HEC}=\frac{888}{175}cm^2\)
Vậy \(S_{AECH}=2.S_{AHE}=\frac{2.888}{175}=\frac{1776}{175}cm^2\)
