由于角paq是直角且在圆上,所以PQ为○的直径.
作线段ah垂直于BC,交PQ于F,易得AH垂直于PQ且F为PQ中点,且F为圆心
则AP = √2/2 * PQ,又因为角DFH = 角ADF + 角DAF = 45度,所以FH = √2/2 * DF = √2/4 * PQ
则AH = AF + FH = (2+√2)/4 * PQ
则AB = (√2+1)/2 * PQ
所以BP = AB - AP = 1/2 * PQ,由对称性,同理得,CP = 1/2 * PQ,所以BP + CQ = PQ
楼上的快乐星雨悠悠说的对,那样证明更加简单.通过对角相等证明BDFP为平行四边形,之后由于FP = FD都是半径,所以BDFP为菱形,所以BP = FP = 1/2PQ