In △ ABC, ∠ ACB = 90 °, M is the midpoint of AB, ∠ PMQ = 90 °, indicating: PQ & sup2; = AP & sup2; + BQ & sup2; P is on AC, q is on BC, I don't have a picture, but I can draw it myself,

In △ ABC, ∠ ACB = 90 °, M is the midpoint of AB, ∠ PMQ = 90 °, indicating: PQ & sup2; = AP & sup2; + BQ & sup2; P is on AC, q is on BC, I don't have a picture, but I can draw it myself,

Using vector to prove. Take C as zero point, two right angle sides as abscissa and ordinate to establish coordinate system (assuming AC as y axis). According to the known relationship, the sitting (0, n), (m, 0), (M / 2, N / 2) scales of a, B and m are obtained. Let P (0, y), q (y, 0). According to vector PM and MQ, we can get MX + NY = (m * m + n * n) / 2pq = under root sign (x * x + y * y)