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有这样一道几何题“如图1,△ABC中,高AD、CG交于F,E为BC的中点。设AD=BC=2a求证EF+DF=c”。这是一个古老的命题。其实,只当△ABC为锐角三角形时为真,其证明可采用几何法或坐标法(证明略)。当△ABC为钝角三角形时我们有下面的结论: 在钝角三角形ABC中,高AD、GC的延长线交于点F,E为BC的中点,若AD=BC=2a,则EF-DF=a,如图2所示。证明:建立如图2所示坐标系,设点F的坐标为(x,y),则点A、B、c的坐标分别为
There is such a geometric problem “Figure 1, △ ABC, high AD, CG pay in F, E is the midpoint of BC. Let AD = BC = 2a to verify EF + DF = c”. This is an old proposition. In fact, it is true only when △ABC is an acute-angled triangle, which can be proved by geometrical method or coordinate method (demonstration omitted). When △ABC is an obtuse triangle, we have the following conclusions: In the obtuse triangle ABC, the extended lines of high AD and GC intersect at point F. E is the midpoint of BC. If AD=BC=2a, then EF-DF= a, as shown in Figure 2. Proof: To establish a coordinate system as shown in Figure 2, set the coordinates of point F to (x,y), then the coordinates of points A, B, and c are