在△ABC中,已知AB=3,AC=6,BC=7,AD是∠BAC平分线.求证:DC=2BD.

问题描述:

在△ABC中,已知AB=3,AC=6,BC=7,AD是∠BAC平分线.求证:DC=2BD.

证明:如图所示,
过点D作DE∥AB交AC于点E.
则∠EDA=∠DAB,又∠DAB=∠EAD,
∴∠EDA=∠EAD,
∴EA=ED.
∵DE∥AB,

CD
DB
CE
EA
CE
ED
AC
AB
=
6
3
=2,
∴CD=2DB.