题以直角三角形ABC的弦AB为边,在直角顶点另侧做正方形ABDE,设BC=a,AC=b,AB=c.试求直角顶点C到正方形中心的距离. 解法1(利用正弦定理)设Q是所作正方形的中心(图1),则∠AQB=90°,于是A、C、B、Q四点共圆,即Q在△ABC的外接圆周上.AB是这外接圆的直径.对△AQC,应用正弦定理有: The question asks that the chord AB of the right-angled triangle ABC is the edge, and the square ABDE is made on the other side of the right-angle vertex. Let BC=a, AC=b, AB=c. Find the distance from the right-angled vertex C to the center of the square. Solution 1 (Using the sine theorem ) Let Q be the center of the square (Fig. 1), then ∠AQB=90°, then A, C, B, Q are all four-points co-circle, that is, Q is on the circumcircle of △ABC. AB is the diameter of this circumcircle For △AQC, apply the sine theorem to: