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The image shows a triangle $$\triangle ABC$$. There are tick marks on sides AC and BC, indicating $$AC = BC$$. So, $$\triangle ABC$$ is an isosceles triangle. The base is AB. We are given $$BC = 8$$.

There is a point M on AC such that BM is drawn. Also, there is a point N on AB such that MN is perpendicular to AB. We are given $$MN = ?$$ and $$AN = ?$$ or some other lengths.

The length 8 is given next to BC, so $$BC = 8$$. Since $$\triangle ABC$$ is isosceles with $$AC = BC$$, we have $$AC = 8$$.

The diagram shows N on AB, and $$MN \perp AB$$. So MN is the altitude from M to AB.

We are given the length 8, which is $$BC$$. Since $$AC = BC$$, $$AC = 8$$.

Without any angles or other lengths, it's impossible to determine the length of MN or AN.

Let's assume there is a missing question, and we need to find something. For example, if we were given the length of AB, or an angle.

If we assume that M is the midpoint of AC, then BM is a median. If $$\triangle ABC$$ is isosceles, the median to the base is also an altitude and an angle bisector. But BM is not the median to the base. M is on AC.

Let's consider the possibility that the question is related to the area of the triangle or the length of MN.

If we assume that $$\triangle ABC$$ is equilateral, then $$AB = BC = AC = 8$$. In this case, M is on AC. If M is the midpoint of AC, then $$AM = MC = 4$$. BM is a median. Since $$\triangle ABC$$ is equilateral, BM is also an altitude, so $$BM \perp AC$$. This contradicts the diagram where M is on AC and BM is drawn as a segment.

Let's assume the diagram is as given, with $$AC = BC = 8$$. N is on AB, $$MN \perp AB$$.

If we consider the area of $$\triangle ABM$$. Area = $$\frac{1}{2} \cdot AB \cdot MN$$.

We need more information to solve this problem. For instance, if we knew the length of AB or the measure of angle C.

Let's assume that the question is to find MN, given some additional information. If we assume that $$\triangle ABC$$ is a right isosceles triangle with $$\angle C = 90^{\circ}$$. Then $$AB = \sqrt{8^2 + 8^2} = 8\sqrt{2}$$.

If M is the midpoint of AC, then $$AM = MC = 4$$. BM is a median. In a right triangle, the median to the hypotenuse is half the hypotenuse. But BM is not the median to the hypotenuse.

Let's assume that the problem is solvable with the given information. The length 8 is associated with BC. There are tick marks on AC and BC, so AC = BC = 8.

There is a point N on AB such that $$MN \perp AB$$. We need to find MN. M is on AC.

This problem cannot be solved without more information, such as angles or the length of AB, or the position of M on AC.