已知：三点A,B,C，A( XA , 177.685393 ),B( XB , 207.438202 ),C( XC, 132.764045 ),
      角ABC=90°，XB-XA=XC-XB,
求计算：XB-XA和XC-XB的长度；

一个参考程序：已知三点坐标，计算边长和角度；如下：
#include<stdio.h>                    
#include<math.h>                    
void main()                    
{                    
double a1,a2,b1,b2,c1,c2,a,b,c,A,B,C;                    
scanf("%lf%lf%lf%lf%lf%lf",&a1,&a2,&b1,&b2,&c1,&c2);                    
a=sqrt((b1-c1)*(b1-c1)+(b2-c2)*(b2-c2));                    
b=sqrt((a1-c1)*(a1-c1)+(a2-c2)*(a2-c2));                    
c=sqrt((a1-b1)*(a1-b1)+(a2-b2)*(a2-b2));                    
A=acos((b*b+c*c-a*a)/(2*b*c))*180/3.14159265;                    
B=acos((a*a+c*c-b*b)/(2*a*c))*180/3.14159265;                    
C=acos((a*a+b*b-c*c)/(2*a*b))*180/3.14159265;                    
printf("A=%lf\nB=%lf\nC=%lf\n",A,B,C);                    
printf("a=%lf\nb=%lf\nc=%lf\n",a,b,c);                    


}                    

对不对我没验证过，仅供参考

#include <stdio.h>
#include <math.h>

int main( void )
{
    double Yab = 207.438202 - 177.685393;
    double Ycb = 207.438202 - 132.764045;
    printf( "%f\n", sqrt(Yab*Ycb) );
}

输出 47.135612

看样子，对啦，

47.135612       177.685393
94.271224       207.438202
141.406836      132.764045
A=57.739175
B=90.000000
C=32.260825
a=88.306260
b=104.426966
c=55.740430
Press any key to continue