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The auxiliary line of the second grade math problem has been done.
Because triangle ABC is an isosceles right triangle, angle ABM= angle ACN;;

Connecting KN, it is known that CK=CN is perpendicular to the X axis, and the known triangle KCN is an isosceles right triangle;

Because triangle ABC is an isosceles right triangle, angle ABN= angle ACM= 135 degrees = angle MAN, AB=AC, indicating that triangle ABN is all equal to triangle ACM and is an isosceles triangle;

So AN=AM, so triangle people are similar to triangle ABN and triangle ACM.

So cm = bn = ab = AC Then: BC+CM = Nb+BC = AB+BC = AC+BC = CN = BM;

Then MN=2AB+BC=2AC+BC, and the value of BM 2+CN 2/Mn 2 remains unchanged.

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