Bài tập theo Tuần toán 8- Tuần 27_ đề 2

Ta có: $DE=AD-AE=17-8=9\left(cm\right)$$\Rightarrow \frac{AB}{DE}=\frac{AE}{DC}\left(Do\frac{6}{9}=\frac{8}{12}\right)\Rightarrow \Delta ABE~\Delta DEC\left(cgc\right)$

$\Rightarrow \angle AEB=\angle DEC\left(1\right);\angle ABE=\angle DEC\left(2\right)$

Từ (1) và (2) suy ra $\angle AEB+\angle DEC={90}^0\Rightarrow \angle BEC={90}^0$

Xét $\Delta ABC$và $\Delta DCB$có: $\angle A=\angle C={90}^0,\frac{AB}{CB}=\frac{BC}{DB}=\frac{2}{3}\Rightarrow \Delta ABC~\Delta DCB(c-g-c)$

$\Rightarrow \angle CBD=\angle ACB,$mà hai góc ở vị trí so le trong $\Rightarrow DB//AC$

Xét $\Delta AHB$và $\Delta CHA$có: $\angle H_1=\angle H_2={90}^0,\angle C=\angle BAH($phụ $\angle HAC)$

$\Rightarrow \Delta AHB~\Delta CHA(g-g)\Rightarrow \frac{AH}{HB}=\frac{CH}{AH}$$\Rightarrow AH=\sqrt{BH.HC}=\sqrt{4.9}=6\left(cm\right)$

$\begin{array}{l}BC=BH+CH=4+9=13\left(cm\right)\Rightarrow AM=\frac{BC}{2}=\mathrm{6,5}\left(cm\right)\\ \Rightarrow HM=\left|BM-BH\right|=\left|\mathrm{6,5}-4\right|=\mathrm{2,5}\left(cm\right)\\ \Rightarrow S_{AHM}=\frac{1}{2}AH.HM=\frac{1}{2}\mathrm{.6,5.2,5}=\frac{65}{8}\left(c{m}^2\right)\end{array}$