Answer:
PM:MQ = 8:3.
Step-by-step explanation:
[tex]\rm \angle B\hat{M}P + 90^{\circ} + \angle C\hat{M}Q = 180^{\circ}[/tex];
[tex]\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^{\circ}[/tex];
[tex]\implies \rm 90^{\circ} - \angle B\hat{M}P = \angle C\hat{M}Q[/tex].
Also,
[tex]\rm \angle B\hat{P}M = 90^{\circ} - \angle B\hat{M}P[/tex] in right triangle PBM.
Thus [tex]\rm \angle{P\hat{B}M} = \angle C\hat{M}Q[/tex].
Additionally [tex]\rm \angle \hat{B} = 90^{\circ} = \angle \hat{C}[/tex].
Therefore [tex]\rm \triangle PBM \sim \triangle MCQ[/tex].
[tex]\rm \displaystyle BC = 2\;MC[/tex] for M is the midpoint of segment BC.
[tex]\rm \displaystyle PB = \frac{4}{3}BC = \frac{8}{3}MC[/tex].
[tex]\rm \triangle PBM \sim \triangle MCQ[/tex] implies that
[tex]\displaystyle \rm PM:MQ = PB:MC = 1:\frac{8}{3} = 8:3[/tex].
