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Trong ΔABC, ta có: AD là đường phân giác của (BAC)
Suy ra: \(\frac{DB}{DC}=\frac{AB}{AC}\)(tính chất đường phân giác)
Mà AB = 15 (cm); AC = 20 (cm)
Nên \(\frac{DB}{DC}=\frac{15}{20}\)
uy ra: \(\frac{DB}{DB+DC}=\frac{15}{15+20}\)(tính chất tỉ lệ thức)
Suy ra: \(\frac{DB}{BC}=\frac{15}{35}\Rightarrow DB=\frac{15}{35}.BC=\frac{15}{35}.25=\frac{75}{7}cm\)
\(\Rightarrow DC=BC-BD=25-\frac{75}{7}=\frac{100}{7}cm\)
b. Kẻ AH ⊥ BC
Ta có: SABD = 1/2 AH.BD; SADC = 1/2 AH.DC
Suy ra :\(\frac{S_{ABD}}{S_{ADC}}=\frac{\frac{1}{2}AH.BD}{\frac{1}{2}AH.DC}=\frac{BD}{DC}\)
Mà\(\frac{DB}{DC}=\frac{15}{20}=\frac{3}{4}\)
\(\Rightarrow\frac{S_{ABD}}{S_{ADC}}=\frac{3}{4}\)
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a) Theo đề bài ta có:
\(\dfrac{AD}{DC}=\dfrac{BA}{BC}=\dfrac{15}{10}=\dfrac{3}{2}\)
\(\dfrac{AD}{AD+DC}=\dfrac{15}{15+10}hay\dfrac{AD}{AC}=\dfrac{15}{25}\)
=> AD = \(\dfrac{15.AC}{25}=\dfrac{15.15}{25}=9\left(cm\right)\)
DC = AC - AD = 15 - 9 = 6 (cm)
Vậy AD = 9cm; DC = 6cm.
b) Vì BD \(\perp\) BE nên BE là đường phân giác của góc ngoài tại đỉnh B.
Áp dụng tính chất đường phân giác của góc ngoài ta có:
\(\dfrac{EC}{EA}=\dfrac{EC}{EC+AC}=\dfrac{BC}{BA}\)
hay \(\dfrac{EC}{EC+15}=\dfrac{10}{15}=\dfrac{2}{3}\)
=> EC = 30 (cm)
Vậy EC = 30cm.
áp dụng định lý pitago vào tam giác vuông ABC:
\(AB^2\)+\(AC^2_{ }=BC^2\)
=>\(AB^2=BC^2-AC^2\)
<=>\(AB^2=6^2-4^2=20=>AB=\sqrt[]{20}\)
ÁP dụng định lý pitago vào tam giác vuông BCD
\(BC^2+DC^2=BD^2=>DC^2=BD^2-BC^2=9^2-6^2=45=>DC=\sqrt[]{45}\)
TA CÓ
\(\dfrac{AB}{CD}=\dfrac{\sqrt[]{20}}{\sqrt[]{45}}=\dfrac{2}{3}\) (1)
\(\dfrac{DC}{BC}=\dfrac{6}{9}=\dfrac{2}{3}\) (2)
TỪ 1 và 2 => \(\Delta ABC\sim\Delta BCD\)
=>\(\widehat{DBC}=\widehat{ACB}\) mà 2 góc này ở vị trí so le trong => BD//AC
Bài giải:
SADEF=\(\dfrac{\left(AD+EF\right).FG}{2}=\dfrac{\left(4+2\right).2}{2}=6\left(cm^2\right)\)
SABCD=\(\dfrac{\left(AD+BC\right).BG}{2}=\dfrac{\left(4+1\right).1}{2}=2,5\left(cm^2\right)\)
=> SABCDEF= SADEF+SABCD= 6+2,5=8,5(cm2)
b) SDEA=\(\dfrac{DE.AE}{2}=\dfrac{4.3}{2}=6\left(cm^2\right)\)
SDCFE=\(\dfrac{\left(DE+CF\right).EF}{2}=\dfrac{\left(4+8\right).4}{2}=24\left(cm^2\right)\)
SCFB=\(\dfrac{CF.FB}{2}=\dfrac{8.6}{2}=24\left(cm^2\right)\)
=> SABCD=SDEA+SDCFE+SCFB=6+24+24=54(cm2)