In $\Delta ABC$,$m \angle A + m \angle C = m \angle B$ and $\overline{BM}$ is an altitude to $\overline{AC}$. If $AM = 16$ and $CM = 9$,find $BM$,$AB$,and $BC$.

(N/A) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$m \angle A + m \angle C + m \angle B = 180^{\circ}$
Since $m \angle A + m \angle C = m \angle B$,we substitute:
$m \angle B + m \angle B = 180^{\circ}$
$2 m \angle B = 180^{\circ}$
$m \angle B = 90^{\circ}$
In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,by the geometric mean theorem:
$BM^2 = AM \times CM$
$BM^2 = 16 \times 9 = 144$
$BM = 12$
Now,using the properties of right triangles:
$AC = AM + CM = 16 + 9 = 25$
$AB^2 = AM \times AC = 16 \times 25 = 400 \implies AB = 20$
$BC^2 = CM \times AC = 9 \times 25 = 225 \implies BC = 15$
Thus,$BM = 12$,$AB = 20$,and $BC = 15$.