证明:根据射影定理有:
AD^2=BD*CD
AB^2=BC*BD
AC^2=BC*CD
1/AC^2=1/(BC*CD)
1/AB^2+1/AC^2
=(AB^2+AC^2)/(AB^2*AC^2)
=BC(BD+CD)/(BC*BD*BC*CD)
=(BC*BC)/(BC*BC*BD*CD)
=1/(BC*CD)
故1/AD^2=1/AB^2+1/AC^2
证明:根据射影定理有:
AD^2=BD*CD
AB^2=BC*BD
AC^2=BC*CD
1/AC^2=1/(BC*CD)
1/AB^2+1/AC^2
=(AB^2+AC^2)/(AB^2*AC^2)
=BC(BD+CD)/(BC*BD*BC*CD)
=(BC*BC)/(BC*BC*BD*CD)
=1/(BC*CD)
故1/AD^2=1/AB^2+1/AC^2