Bài 1:
1: Xét ΔBDM và ΔBAC có
\(\hat{BDM}=\hat{BAC}\) (hai góc đồng vị, DM//AC)
\(\hat{DBM}\) chung
Do đó: ΔBDM~ΔBAC
Xét ΔBAC và ΔMEC có
\(\hat{BCA}\) chung
\(\hat{BAC}=\hat{MEC}\) (hai góc đồng vị, ME//AB)
Do đó: ΔBAC~ΔMEC
mà ΔBDM~ΔBAC
nên ΔBDM~ΔMEC
Xét ΔBAC có DM//AC
nên \(\frac{DM}{AC}=\frac{BM}{BC}=\frac{BD}{BA}\)
Xét ΔBAC có ME//AB
nên \(\frac{EM}{AB}=\frac{CM}{CB}=\frac{CE}{CA}\)
\(\frac{DM}{AC}+\frac{EM}{AB}=\frac{BM}{BC}+\frac{CM}{BC}=\frac{BC}{BC}=1\)
2: ΔBDM~ΔMEC
=>\(\frac{S_{BDM}}{S_{MEC}}=\left(\frac{BM}{MC}\right)^2\)
=>\(\left(\frac{BM}{MC}\right)^2=\frac94=\left(\frac32\right)^2\)
=>\(\frac{BM}{MC}=\frac32\)
=>\(\frac{BM}{BC}=\frac35\)
ΔBDM~ΔBAC
=>\(\frac{S_{BDM}}{S_{BAC}}=\left(\frac{BM}{BC}\right)^2=\frac{9}{25}\)
=>\(S_{BAC}=25\left(\operatorname{cm}^2\right)\)
Ta có: \(S_{BDM}+S_{ADME}+S_{EMC}=S_{ABC}\)
=>\(S_{ADME}=25-4-9=12\left(\operatorname{cm}^2\right)\)
