△ABC的三边AB、BC、CA长分别是20、30、40，其三条角平分线将△ABC分为三个三角形，则S△ABO︰S△BCO︰S△CAO等于（ ）
A．1︰1︰1 B．1︰2︰3 C．2︰3︰4 D．3︰4︰5
答案：C
题目分析：首先过点O，作OD⊥AB于D，作OE⊥AC于E，作OF⊥BC于F，由点O是△ABC内角平分线的交点，根据角平分线的性质，即可得OD=OE=OF，继而可得
：
：
=AB：BC：CA，则可求得答案．
过点O，作OD⊥AB于D，作OE⊥AC于E，作OF⊥BC于F，
∵点O是△ABC内角平分线的交点，
∴OD=OE=OF，
∴
，
，
，
∵AB=20，BC=30，CA=40，
∴：： =AB：BC：CA =20：30：40=2：3：4．
故选C．
点评：解答本题的关键是掌握角平分线上的点到角的两边的距离相等，三角形三个内角平分线的交点到三边的距离相等。
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