point z is the circumcenter of δlmn. which must be true?
○ $overline{an} cong overline{lb}$
○ $overline{nc} cong overline{bz}$
○ $angle bmz cong angle cmz$
○ $angle zbm cong angle zcm$

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# Brief Explanations:
The circumcenter $ Z $ of a triangle is the intersection of the perpendicular bisectors of the triangle's sides. So, $ ZA \perp LN $, $ ZB \perp LM $, and $ ZC \perp MN $, and $ A $, $ B $, $ C $ are midpoints (since perpendicular bisectors bisect the sides).

- For $ \overline{AN} \cong \overline{LB} $: $ AN=\frac{1}{2}LN $ and $ LB = \frac{1}{2}LM $. There's no info $ LN = LM $, so not necessarily true.
- For $ \overline{NC} \cong \overline{BZ} $: $ NC=\frac{1}{2}MN $, $ BZ $ is a segment from $ Z $ to $ B $ (on $ LM $). No reason for these to be congruent.
- For $ \angle BMZ \cong \angle CMZ $: $ ZC \perp MN $ and $ ZB \perp LM $, but $ Z $ is circumcenter, not necessarily angle bisector of $ \angle LMN $ unless triangle is isoceles with $ LM = LN $, which isn't given.
- For $ \angle ZBM \cong \angle ZCM $: $ \triangle ZBM $ and $ \triangle ZCM $: $ \angle ZBM=\angle ZCM = 90^\circ $ (since $ ZB \perp LM $, $ ZC \perp MN $), $ ZM = ZM $ (common hypotenuse), and $ ZB = ZC $ (distances from circumcenter to sides? Wait, no—circumcenter is equidistant from vertices, but here $ ZB $ and $ ZC $ are distances from $ Z $ to sides? Wait, no, $ Z $ is circumcenter, so $ ZL = ZM = ZN $ (distances to vertices). But $ ZB $ and $ ZC $ are perpendiculars to sides, so in right triangles $ ZBM $ and $ ZCM $: $ \angle ZBM=\angle ZCM = 90^\circ $, $ ZM = ZM $, and $ ZB = ZC $? Wait, no, actually, since $ Z $ is circumcenter, the perpendicular bisectors: $ B $ is midpoint of $ LM $, $ C $ is midpoint of $ MN $? Wait, no, the perpendicular bisector of $ LM $ passes through $ Z $ and $ B $ (with $ ZB \perp LM $), so $ B $ is midpoint of $ LM $, so $ LB = BM $. Similarly, $ C $ is midpoint of $ MN $, so $ NC = CM $. Now, in $ \triangle ZBM $ and $ \triangle ZCM $: $ \angle ZBM = \angle ZCM = 90^\circ $, $ ZM = ZM $ (hypotenuse), and $ BM $ and $ CM $? Wait, no, maybe better: $ ZB $ and $ ZC $ are both perpendicular to sides, but actually, since $ Z $ is circumcenter, the distances from $ Z $ to the sides (the lengths $ ZB $, $ ZC $, $ ZA $) are not necessarily equal (that's the inradius, for incenter). Wait, I made a mistake. Circumcenter is equidistant from vertices, incenter is equidistant from sides. So $ ZL = ZM = ZN $, but $ ZB $, $ ZC $, $ ZA $ are not necessarily equal. So let's re-examine the angles. $ \angle ZBM $ and $ \angle ZCM $: both are right angles? Wait, $ ZB \perp LM $, so $ \angle ZBM = 90^\circ $, $ ZC \perp MN $, so $ \angle ZCM = 90^\circ $. Wait, no, $ ZB $ is perpendicular to $ LM $, so $ \angle ZBM = 90^\circ $, $ ZC $ is perpendicular to $ MN $, so $ \angle ZCM = 90^\circ $. Wait, but that would mean $ \angle ZBM \cong \angle ZCM $ because both are right angles (90 degrees). Wait, is that correct? Let's check the diagram: $ ZB $ is perpendicular to $ LM $ (right angle at $ B $), $ ZC $ is perpendicular to $ MN $ (right angle at $ C $). So $ \angle ZBM = 90^\circ $ and $ \angle ZCM = 90^\circ $, so they are congruent. Wait, but let's check the other options again.

Wait, the circumcenter is the intersection of perpendicular bisectors, so $ B $ is the midpoint of $ LM $ (since $ ZB \perp LM $ and $ Z $ is on the perpendicular bisector), so $ LB = BM $. Similarly, $ C $ is midpoint of $ MN $, so $ NC = CM $.

Now, $ \angle BMZ \cong \angle CMZ $: for this, $ ZM $ would have to be the angle bisector, but $ Z $ is circumcenter, not necessarily angle bisector unless triangle is isoceles with $ LM = MN $, which isn't given.

$ \overline{AN} \cong \overline{LB} $: $ AN = \frac{1}{2}LN $, $ LB = \frac{1}{2}LM $. No info $ LN = LM $, so not true.

$ \overline{NC} \cong \overline{BZ} $: $ NC = \frac{1}{2}MN $, $ BZ $ is a segment from $ Z $ to $ B $, no reason to be equal.

$ \angle ZBM \cong \angle ZCM $: both are right angles (90°), so they are congruent. So this must be true.
# Answer: $ \boldsymbol{\angle ZBM \cong \angle ZCM} $ (the last option)

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