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本文介绍经过三角形内一点的三条直线的一个性质,以下分为几种情形阐述,与读者共赏.情形1如图1,P是△ABC内一点,过点P分别引直线DE∥BC,FG∥AC,HK∥AB,则PD PE·PF PG·PH PK=1.证明作射线AP、BP、CP交对边分别为点A′、B′、C′,
This article describes a triangle through a triangle within a point of a nature, the following is divided into several situations elaborated, and the audience together. Scenario 1 as shown in Figure 1, P is a point within △ ABC, point P respectively lead straight line DE∥BC, FG ∥AC, HK∥AB, then PD PE · PF PG · PH PK = 1. It is proved that the rays AP, BP and CP cross the points A ’, B’, C ’