To explore other points secant triangle _11215

By Laura Jenkins,2014-11-29 11:37
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To explore other points secant triangle _11215

To explore other points secant triangle

     How can an area of a triangle divided into two equal parts, is that we have well known problems, as long as the triangle along the median line, you can put a triangle divided into two parts of equal size,

    and many students believe that such a split line only three, However, such

    a division line in the end how many do?

     Question 1: Please use a straight line to an area of ?

    ABC divided into two equal parts.

     Solution: Take the midpoint of BC, denoted by point D, links to AD, the AD where the straight line to ? ABC is divided into two

    parts of equal size.

     As we all know, so that a total of three separate lines, namely ? ABC through the middle of three straight, the area of ? ABC

    can be divided into equal two parts. In addition to these three, there are many, and for the edge of ? ABC at any point, can be found one after this point and the triangle area is split between lines.

     Question 2: Point E is the edge of ? ABC, AB at any

    point, and AE ? BE, Guo point E seeking to make a straight line, the area of ? ABC is divided into two equal parts.

     Solution: Figure 2, the midpoint of AB taking D, Link CD, over point D for DF ? CE, cross BC at point F, then the straight line EF is the desired cutting line.

     Proof: Let CD, EF intersect at point P

     ? point D is the midpoint of AB

     ? AD = BD ? S ? CAD = S ? CBD

     ? S quadrilateral CAEP + S ? PED = S quadrilateral

    DPFB + S ? PCF

     Also ? DF ? CE ? S ? FED = S ? DCF (with the

    bottom contour)

     Namely: S ? PED = S ? PCF

     ? S quadrilateral CAEP = S quadrilateral DPFB

     ? S quadrilateral CAEP + SPCF = S quadrilateral DPFB + S

    ? PED

     Namely, S quadrilateral AEFC = S ? EBF

     From this, the triangle area to split the line there are numerous articles and

     Besides, after the edge of any of a straight line, the use

    of the special nature of the diagonal ladder, it is easy to make such a split line.

     So, these dividing lines, will pay a particular a specific point?

     As you know, the triangle's three middle regarded area of a triangle divided into two equal parts, while the three middle pay on its center of gravity, if the split lines intersect at one point, then this

point must be the focus of a triangle.

     Question 3: It is known: in Figure 3, in ? ABC,, G is

    the center of ? ABC, over point of G to EF ? BC AC AB at points E, AC

    AC at point F, verify: S ? AEF = S ? ABC .

     Prove: Extension AG, cross BC at point D

     ? point G is the center of ? ABC

     ? AG: AD = 2:3

     Also ? EF ? BC, ? ? AEF ? ? ABC

     Available by the question: over the edge AB point E, a straight line through the center of gravity G's, EF the triangle area is divided into two parts, 4:5, a straight line EF is not a triangle, etc.

    secant points. According to Question 2, you can find a cross point E to the triangle area is split in a straight line, this line will however focus on G.

     To sum up, triangle, etc. There are countless points of

    the secant, and the edge of any given point, can make such points the corresponding secant, and only one, all the dividing lines do not intersect at the triangle center of gravity. Reposted elsewhere in the paper for free download http://

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