In rhombus $ABCD$,$AB = 25$ and $AC = 14$. Find $BD$.

In $\Delta ABC$,$m \angle A = m \angle B + m \angle C$ and $\overline{AM}$ is an altitude. If $AM = \sqrt{12}$ and $BC = 8$,find $BM$.

In the figure,$AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that:
$AB^{2} = AD^{2} - BC \cdot DM + \left(\frac{BC}{2}\right)^{2}$

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $AD = 12$ and the perimeter of $\Delta ABC$ is $36$,find the area of $\Delta ABC$.

In rhombus $ABCD$,$AC > BD$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $AM + DM = 17$ and $AB = 13$,find $BD$.

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