Triangle ABC𝐴𝐵𝐶 has a right angle at A𝐴, and side AC𝐴𝐶 is twice as long as AB𝐴𝐵. Also MN𝑀𝑁 is the perpendicular bisector of BC𝐵𝐶, as shown.In area, what fraction of △ABC△𝐴𝐵𝐶 is △CMN△𝐶𝑀𝑁?

To solve this problem, we need to understand a few properties of triangles.

1. The area of a triangle is given by 1/2 * base * height.
2. In a right triangle, the two sides forming the right angle are considered the base and the height.
3. The perpendicular bisector of the hypotenuse of a right triangle divides the triangle into two smaller triangles, each of which is similar to the original triangle and to each other.

Given that triangle ABC is a right triangle with a right angle at A, and AC is twice as long as AB, we can assign arbitrary values to these lengths for simplicity. Let's say AB = x and AC = 2x. Therefore, BC (the hypotenuse) = sqrt[(2x)^2 + x^2] = sqrt[5x^2].

MN, the perpendicular bisector of BC, divides BC into two equal segments. So, BM = CN = 1/2 * sqrt[5x^2] = sqrt[5x^2]/2.

Now, let's find the areas of triangle ABC and triangle CMN.

Area of triangle ABC = 1/2 * AB * AC = 1/2 * x * 2x = x^