Giải thích các bước giải:

Bài 2:

a.Ta có $BM, CN$ là phân giác $\hat B, \hat C$

$\to \dfrac{MA}{MC}=\dfrac{BA}{BC}=\dfrac{CA}{CB}=\dfrac{NA}{NB}$

$\to MN//CB$

b.Xét $\Delta ANC,\Delta AMB$ có:

Chung $\hat A$

$\widehat{ACN}=\dfrac12\hat C=\dfrac12\hat B=\widehat{ABM}$

$\to\Delta ANC\sim\Delta AMB(g.g)$

c.Ta có:

$\dfrac{MA}{MC}=\dfrac{BA}{BC}=\dfrac56$

$\to \dfrac{MA}{MA+MC}=\dfrac5{5+6}$

$\to \dfrac{MA}{AC}=\dfrac5{11}$

$\to MA=\dfrac5{11}AC$

$\to AM=\dfrac{25}{11}$

Lại có $MN//BC$

$\to \dfrac{MN}{BC}=\dfrac{AM}{AC}=\dfrac5{11}$

$\to MN=\dfrac5{11}BC$

$\to MN=\dfrac{30}{11}$

d.Gọi $AH\perp BC\to H$ là trung điểm $BC\to HB=HC=\dfrac12CB=3$

$\to AH=\sqrt{AB^2-BH^2}=4$

$\to  S_{ABC}=\dfrac12AH.BC=12$

Ta có $MN//BC$

$\to \widehat{AMN}=\widehat{ACB},\widehat{ANM}=\widehat{ABC}$

$\to\Delta ANM\sim\Delta ABC(g.g)$

$\to \dfrac{S_{ANM}}{S_{ABC}}=(\dfrac{AM}{AC})^2=\dfrac{25}{121}$

$\to S_{AMN}=\dfrac{300}{121}$

Bài 3:

a.Ta có $\Delta ABC$ vuông tại $A\to BC=\sqrt{AB^2+AC^2}=10$

Mà $BD$ là phân giác $\hat B$

$\to \dfrac{DA}{DC}=\dfrac{BA}{BC}=\dfrac35$

$\to \dfrac{AD}{AD+DC}=\dfrac3{3+5}$

$\to \dfrac{AD}{AC}=\dfrac38$

$\to AD=\dfrac38AC=3$

$\to CD=AC-AD=5$

b.Xét $\Delta ABH,\Delta ABC$ có:

Chung $\hat B$

$\widehat{AHB}=\widehat{BAC}(=90^o)$

$\to\Delta BHA\sim\Delta BAC(g.g)$
$\to \dfrac{BH}{BA}=\dfrac{BA}{BC}$

$\to AB^2=BH.BC$

c.Xét $\Delta ABI,\Delta CDB$ có:

$\widehat{ABI}=\widehat{DBC}$ vì $BD$ là phân giác $\hat B$
$\widehat{BAI}=\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}=\widehat{DCB}$

$\to\Delta ABI\sim\Delta CBD(g.g)$

d.Ta có $BI, BD$ là phân giác $\hat B$

$\to \dfrac{IH}{IA}=\dfrac{BH}{BA}=\dfrac{BA}{BC}=\dfrac{DA}{DC}$

$\to IH.DC=IA.DA$