Let $[ABCD]$ be a square, $M$ a point on the segment $[AD]$, and $N$ a point on the segment $[DC]$ such that $B\hat{M}A = N\hat{M}D = 60^{\circ}$. Calculate $M\hat{B}N$.

We are given a square $ ABCD $ with points $ M $ and $ N $ lying on segments $ AD $ and $ DC $, respectively. The angles $ B\hat{M}A = N\hat{M}D = 60^\circ $, and we are tasked with finding $ M\hat{B}N $.

### Step 1: Analyze the Geometry

Let's start by considering the geometric properties of the square $ ABCD $. In a square, all sides are equal, and all angles are right angles. Suppose the side length of the square is $ s $, so:

\[
AB = BC = CD = DA = s
\]
Also, since it is a square, the diagonals are equal and bisect each other at $ 45^\circ $.

### Step 2: Placement of Points $ M $ and $ N $

- Let $ M $ be a point on segment $ AD $, and let the distance from $ A $ to $ M $ be $ x $, so that $ AM = x $ and $ MD = s - x $.
- Let $ N $ be a point on segment $ DC $, and let the distance from $ D $ to $ N $ be $ y $, so that $ DN = y $ and $ NC = s - y $.

### Step 3: Angles at $ M $

We are given two angles:
- $ B\hat{M}A = 60^\circ $
- $ N\hat{M}D = 60^\circ $

This means that the configuration around $ M $ forms two equilateral triangles involving points $ B, A, M $ and $ N, D, M $.

### Step 4: Use of Symmetry

The angles $ B\hat{M}A $ and $ N\hat{M}D $ being both $ 60^\circ $ suggests symmetry in the figure. Since these angles indicate equilateral triangle-like structures, and $ M $ is on segment $ AD $ while $ N $ is on segment $ DC $, it follows that the overall shape will exhibit rotational symmetry about point $ M $.

### Step 5: Calculating $ M\hat{B}N $

The key observation is that because of the symmetry and the $ 60^\circ $ angles at $ M $, the angle $ M\hat{B}N $ is part of a cyclic quadrilateral formed by $ M, B, N $. In such a configuration, the external angle (formed by extending $ BM $ and $ MN $) will sum to $ 360^\circ $ with the interior angles at $ M $ and $ N $.

Therefore, since both $ B\hat{M}A $ and $ N\hat{M}D $ are $ 60^\circ $, the angle $ M\hat{B}N $ will be:

\[
M\hat{B}N = 180^\circ - (60^\circ + 60^\circ) = 60^\circ
\]

Thus, the required angle $ M\hat{B}N = 60^\circ $.

### Final Answer:
\[
M\hat{B}N = 60^\circ
\]

Would you like further details or have any questions?

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Here are 5 related questions to deepen understanding:
1. How can we generalize the result if $ B\hat{M}A = N\hat{M}D = \theta $ instead of $ 60^\circ $?
2. What is the significance of symmetry in problems involving squares and specific angle conditions?
3. How would the problem change if the figure was a rectangle instead of a square?
4. What geometric properties of squares make angle-chasing a useful technique?
5. Can this problem be solved using coordinate geometry or vectors for verification?

**Tip**: In problems involving angles and symmetry, look for possible rotational or reflective symmetries to simplify the calculation.

This problem explores geometric properties in a square where two angles at specific points are given as 60°. Learn how to apply symmetry, angle-chasing techniques, and the properties of cyclic quadrilaterals to calculate the unknown angle M̂BN. Suitable for Grades 9-12.

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