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.
答
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
=2,6 3
∴CD=2DB.