Question

直角三角形ABC,C为垂足,过点C做CD垂直AB,求证CD的平方=AD*DB,BC的平方=BD*BA,CA的平方=AD*

Answer 1

CD^2=AC^2-AD^2

CD^2=BC^2-DB^2

2CD^2=AC^2-AD^2+BC^2-DB^2

=AC^2+BC^2-AD^2-DB^2

=AB^2-AD^2-BD^2

=AD^2+BD^2+2AD*DB-AD^2-DB^2

=2AD*DB

所以CD^2=AD*DB

BC^2=BD^2+CD^2

=BD^2+AD*DB

=BD*(BD+AD)

=BD*AB

AC^2=AD^2+CD^2

=AD^2+AD*DB

=AD*(BD+AD)

=AD*AB