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The image shows a figure with points A, B, D, and M. We are given that $$AM = MB = AB$$ and $$DE = 4$$. There are right angle symbols at D and E, suggesting perpendicular lines. It appears that $$\triangle ABD$$ and $$\triangle ABM$$ are involved.

The condition $$AM = MB = AB$$ implies that $$\triangle ABM$$ is an equilateral triangle. This means all angles in $$\triangle ABM$$ are $$60^{\circ}$$.

However, the diagram shows a right angle at D, with AD perpendicular to AB. Also, there is a point E with a right angle at E, suggesting ME is perpendicular to AB or AD. The information $$DE = 4$$ is also given.

Given the labels and right angles, it is likely that D is on AM and E is on BM or AB.

If $$\triangle ABM$$ is equilateral, then $$\angle BAM = \angle ABM = \angle AMB = 60^{\circ}$$.

If AD is perpendicular to AB, then $$\angle DAB = 90^{\circ}$$. This contradicts $$\angle BAM = 60^{\circ}$$ unless M lies on AD extended, which doesn't seem to be the case.

There seems to be a contradiction in the provided information and diagram, or some information is missing or misinterpreted.

Assuming the diagram intends for $$\triangle ABD$$ to be a right-angled triangle at D, and ME is perpendicular to AB at E. If $$AM = MB = AB$$, then $$\triangle ABM$$ is equilateral. Let $$AB = s$$. Then $$AM = MB = s$$.

If AD is perpendicular to AB, then $$\angle DAB = 90^{\circ}$$. This would make $$\triangle ABD$$ a right-angled triangle. But D is on AM.

Let's re-examine the diagram. It seems A, D, M are collinear, and A, E, B are collinear, and B, M are connected. Also, BD is drawn, and ME is drawn.

If $$AM=MB=AB$$, then $$\triangle ABM$$ is equilateral. $$\angle BAM = 60^{\circ}$$.

However, the diagram shows $$\angle ADB = 90^{\circ}$$ and $$\angle MEB = 90^{\circ}$$. This implies D is on AM and E is on AB.

If $$\angle ADB = 90^{\circ}$$, then in $$\triangle ABD$$, $$AD = AB \cos(60^{\circ}) = s/2$$ and $$BD = AB \sin(60^{\circ}) = s\sqrt{3}/2$$. Since D is on AM and $$AM=s$$, $$DM = AM - AD = s - s/2 = s/2$$. So AD=DM, D is the midpoint of AM.

If $$\angle MEB = 90^{\circ}$$, then ME is the altitude from M to AB. In an equilateral triangle, the altitude is also the median. So E is the midpoint of AB. $$AE = EB = s/2$$.

We are given $$DE = 4$$.

Consider the coordinates. Let A = (0, 0), B = (s, 0). Then M = $$(s/2, s\sqrt{3}/2)$$.

D is the midpoint of AM. $$D = ( (0+s/2)/2, (0+s\sqrt{3}/2)/2 ) = (s/4, s\sqrt{3}/4)$$.

E is the midpoint of AB. $$E = (s/2, 0)$$.

Now calculate the distance DE:

• $$DE^2 = (x_D - x_E)^2 + (y_D - y_E)^2$$
• $$DE^2 = (s/4 - s/2)^2 + (s\sqrt{3}/4 - 0)^2$$
• $$DE^2 = (-s/4)^2 + (s\sqrt{3}/4)^2$$
• $$DE^2 = s^2/16 + 3s^2/16$$
• $$DE^2 = 4s^2/16 = s^2/4$$
• $$DE = \sqrt{s^2/4} = s/2$$

We are given $$DE = 4$$. So, $$s/2 = 4$$, which means $$s = 8$$.

Therefore, $$AB = AM = MB = 8$$.

Final Answer: The side length of the equilateral triangle ABM is 8.