In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AB} \cong \overline{BC}$. Find the ratio $AB : AC$.

As shown in the adjoining diagram,$\angle B$ is an obtuse angle and $\overline{AM}$ is an altitude in $\Delta ABC$. Prove that $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.

It is given that $\triangle ABC \sim \triangle DFE$,$\angle A = 30^{\circ}$,$\angle C = 50^{\circ}$,$AB = 5 \, cm$,$AC = 8 \, cm$ and $DF = 7.5 \, cm$. Then,the following is true:

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 3x - 1$,$MB = 2x + 1$,$AN = 3x + 1$ and $NC = 5x + 1$,find the value of $x$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. If $BD = 2 \sqrt{30}$ and $CD = 6$,then $AC = \ldots$

In $\Delta ABC$,$m\angle B = 90^\circ$ and $\overline{BM}$ is an altitude. If $AB = 8$ and $BC = 6$,find $AM$,$BM$,and $CM$.