1. No triângulo $\,ABC\,$, construimos $\, \overleftrightarrow{MC}\,//\, \overleftrightarrow{AS}\,\longrightarrow\;$ pelo teorema fundamental do paralelismo temos
$\,\hat{M}\,=\,B\hat{A}S\,=\,\alpha\,$ (ângulos correspondentes)
$\,\hat{C}\,=\,C\hat{A}S\,=\,\alpha\,$ (alternos internos)
Se $\,\hat{M}\,=\,\hat{C}\,=\,\alpha\,\therefore\,\,\triangle ACM\,$ é isósceles com $\,\boxed{\,\overline{AC}\,\cong\,\overline{AM}\,}$
2. Pelo
Teorema de Tales:
$\phantom{X}\dfrac{AB}{AM}\,=\,\dfrac{BS}{CS}\phantom{X}$, mas $\,\overline{AC}\,\cong\,\overline{AM}\,$ então: $\phantom{X}\dfrac{AB}{AC}\,=\,\dfrac{BS}{CS}\phantom{X}$
c.q.d.