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(C) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Given $m \angle A = m \angle B + m \angle C$,substitutin

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$2$ or $6$

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(C) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Given $m \angle A = m \angle B + m \angle C$,substituting this into the sum gives $m \angle A + m \angle A = 180^{\circ}$,so $2m \angle A = 180^{\circ}$,which means $m \angle A = 90^{\circ}$.Since $\overline{AM}$ is an altitude to $\overline{BC}$,$\Delta ABM$ and $\Delta ACM$ are right-angled triangles at $M$.Let $BM = x$. Then $MC = 8 - x$.In $\Delta ABM$,$AB^2 = AM^2 + BM^2 = 12 + x^2$.In $\Delta ACM$,$AC^2 = AM^2 + MC^2 = 12 + (8 - x)^2$.In $\Delta ABC$,by the Pythagorean theorem,$AB^2 + AC^2 = BC^2$,so $(12 + x^2) + (12 + (8 - x)^2) = 8^2$.$24 + x^2 + 64 - 16x + x^2 = 64$.$2x^2 - 16x + 24 = 0$.$x^2 - 8x + 12 = 0$.$(x - 6)(x - 2) = 0$.Thus,$x = 6$ or $x = 2$.

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 $BM$.

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