Đáp án:

a) $\triangle ANC\backsim\triangle AMB$

b) $MN//BC, AM=\dfrac{25}{11}cm$

c) $S_{\triangle AMN}=\dfrac{300}{121}cm^2$

Giải thích các bước giải:

a)

$\triangle ABC$ cân tại A (gt)

$\to\widehat{ABC}=\widehat{ACB}$ (2 góc ở đáy)

$\to\widehat{ABM}=\widehat{CBM}=\widehat{ACN}=\widehat{BCN}$

Xét $\triangle ANC$ và $\triangle AMB$:

$\widehat{ACN}=\widehat{ABM}$ (cmt)

$\widehat{A}$: chung

$\to\triangle ANC\backsim\triangle AMB$ (g.g)

b)

$\triangle ABC$ có đường phân giác BM (gt)

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

$\triangle ABC$ có đường phân giác CN (gt)

$\to\dfrac{NA}{NB}=\dfrac{CA}{CB}$

Ta có: $BA=CA$ (gt)

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

Xét $\triangle ABC$:

$\dfrac{MA}{MC}=\dfrac{NA}{NB}$ (cmt)

$\to MN//BC$ (định lý Talet đảo)

Ta có: $\dfrac{MA}{MC}=\dfrac{BA}{BC}=\dfrac{5}{6}$

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

$MA+MC=AC=5(cm)\\\to\dfrac{5}{6}MC+MC=5\\\to 11MC=30\\\to MC=\dfrac{30}{11}(cm)\to MA=\dfrac{5}{6}.\dfrac{30}{11}=\dfrac{25}{11}(cm)$

c)

Xét $\triangle AMN$ và $\triangle ACB$:

$\widehat{AMN}=\widehat{ACB}$ (đồng vị)

$\widehat{A}$: chung

$\to\triangle AMN\backsim\triangle ACB$ (g.g)

$\to k=\dfrac{AM}{AC}$

Ta có: $AM+MC=AC$

$\to AM+\dfrac{6}{5}AM=AC\\\to 11AM=5AC\\\to AM=\dfrac{5}{11}AC\\\to k=\dfrac{\dfrac{5}{11}AC}{AC}=\dfrac{5}{11}$

$\to\dfrac{S_{\triangle AMN}}{S_{\triangle ACB}}=k^2=\left(\dfrac{5}{11}\right)^2=\dfrac{25}{121}\\\to S_{\triangle AMN}=\dfrac{25}{121}.12=\dfrac{300}{121}(cm^2)$