Giải thích các bước giải:

Bài 3:

a.Ta có $AD$ là phân giác $\widehat{BAC}$

$\to \dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac9{10}$

$\to \dfrac{BD}{BD+DC}=\dfrac9{10+9}$

$\to \dfrac{BD}{BC}=\dfrac9{19}$

$\to BD=\dfrac9{19}BC=\dfrac{63}{19}$

$\to CD=BC-BD=\dfrac{70}{19}$

b.Ta có $DH//AB$

$\to \dfrac{DH}{AB}=\dfrac{CD}{BC}=\dfrac{10}{19}\to DH=\dfrac{10}{19}AB=\dfrac{90}{19}$

c.Xét $\Delta CDH, \Delta CAB$ có:

Chung $\hat C$

$\widehat{CDH}=\widehat{CBA}$ vì $DH//AB$

$\to\Delta CDH\sim\Delta CBA(g.g)$

Bài 4:

a.Ta có $\Delta ABC$ vuông tại $A$

$\to BC=\sqrt{AB^2+AC^2}=15$

Mà $AD$ là phân giác $\hat A$

$\to \dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac34$

$\to\dfrac{DB}{DB+DC}=\dfrac3{3+4}$

$\to\dfrac{DB}{CB}=\dfrac37$

$\to BD=\dfrac37BC=\dfrac{45}{7}$

$\to CD=BC-BD=\dfrac{60}{7}$

Mà $DE\perp AC, AB\perp AC\to DE//AB$

$\to \dfrac{DE}{AB}=\dfrac{CD}{CB}=\dfrac47$

$\to DE=\dfrac47AB=\dfrac{36}{7}$

b.Ta có $S_{ABC}=\dfrac12AB\cdot AD=54$

Lại có:

$\dfrac{S_{ABD}}{S_{ABC}}=\dfrac{BD}{BC}=\dfrac37$

$\to S_{ABD}=\dfrac37S_{ABC}=\dfrac{162}{7}$

$\to S_{ACD}=S_{ABC}-S_{ABD}=\dfrac{216}{7}$

Bài 5:

a.Xét $\Delta ABD,\Delta BCD$ có:

$\widehat{DAB}=\widehat{DBC}$

$\widehat{ABD}=\widehat{BDC}$ vì $AB//CD$

$\to\Delta ABD\sim\Delta BDC(g.g)$

b.Từ câu a

$\to \dfrac{AB}{BD}=\dfrac{BD}{CD}=\dfrac{AD}{BC}$

$\to \dfrac{2.5}{BD}=\dfrac{BD}{CD}=\dfrac{3.5}{5}$

$\to BD=\dfrac{27}{5}\to CD=\dfrac{250}{49}$