What is the same method in plane geometry?

The same law: On the premise of conforming to the same law, a method of proving the inverse proposition of the original proposition instead of proving it is called the same law. The general steps to prove with the same method are:

1) do not start from the known conditions, but make a graph that conforms to the characteristics of the conclusion; (2) prove that the resulting figure meets the known conditions; (3) Extrapolate the figure that has been made is known to the present.

Example: It is known that n is a point on the edge of the BC of a square ABCD, extending Ba to m, so that Am=CN, as de mn, and E as a vertical foot. Verify: Vertical foot on the e** segment AC.

Proof: Let the intersection of AC and MN be the point F, connecting af, dm, and dn

Obviously, rt mad rt ncd is easily proven, so dm=dn, mda= ndc

So mdn= mda+ adn= ndc+ adn= adc=90°, so dmn is an isosceles right triangle, so dmf=45°, and daf=45°, so dmf= daf, so the quadrilateral mafd is a circle with a quadrilateral, so mfd= mad=90°, i.e. df mn, and de mn, so it can be seen that df and de are the same straight line, and the point f and the point e are actually the same point ( There is and only one line perpendicular to the known straight line at a point outside the line), and f is the intersection of ac and mn, and of course on ac, which proves that de mn's perpendicular foot e is on ac.

Note: It is not easy to use the direct evidence method for this question, and the indirect evidence method (same method, unity method, etc.) can be used instead

When e --.When the arc dg of circle A is s3-->0, s1 >semicircle》s2+s4, the proposition is not true.

cd fg, by the intersecting string theorem, ec*ed=ef*eg, s ecf=(1 2)ec*ef, s ecg=(1 2)ec*eg, s edf=(1 2)ed*ef, s edg=(1 2)ed*eg, so s ecf*s edg=s ecg*s edf.

Cannot be converted to the equation of s1, s2, s3, s4.

Actually, the plane geometry problem? It's all up to the mark. It's like in. The construction site is the main body. Like the tread of that staircase. Is there a geometric problem with the combination? And it's like the Pythagorean theorem they say.

Proof: Take the point q on the pf so that d, q, c, and p are all round.

Connect DQ, QC, extend QC, PB to point G, extend DQ, AP to point H, connect EG, EH

Proband: h, e, g three-point collinear.

Straight line qcg truncated triangle FBP

By Menelaus's theorem:

gbpg ×

cfbc ×

qpfq =1

In the same way: from the straight line dqh truncated dfap:

hpah ×

qfpq ×

dafd =1

Truncated dfab by straight line DCE:

eabe ×

cbfc ×

dfad =1

Multiplication of three forms: GBPG

cfbc ×

qpfq ×

hpah ×

qfpq ×

dafd ×

eabe ×

cbfc ×

dfad =1

Simplified: hpah

gbpg ×

eabe =1

From Menelaus's inverse theorem: h, e, g three points collinear, d, q, c, p four points are conspecific.

hqc= dpc=180- apb= gph p,q,h, g four-point contour.

PCD= PQD= PGE P,C,E, G DPF= DCQ= ECG= EPB

This is a geometry problem in the 2010 Southeast Mathematics Competition, as shown in the figure below

Two questions at a time, not easy, 11 questions and 12 questions.

Analysis: For (1) the judgment can be based on the definition of a straight line on a different surface, for (2) the judgment can be made according to the judgment theorem that the line and surface are perpendicular, for (3) the judgment can be made according to the judgment theorem that the surface is parallel, and for (4) it is possible to list all possible

Answer: Solution: (1) m, l = a, point a m, then l and m are not coplanar, according to the definition of the straight line of the opposite surface, it can be known that it is correct;

2) l and m are straight lines on different planes, l , m, and n l, n m, then n is correct according to the determination theorem that the line and plane are perpendicular;

3) If l , m, then l and m are parallel, intersecting, and different planes, so it is incorrect;

4) If l, m, l m = point a, l , m , then the correct is known according to the determination theorem that the surface is parallel;

So the answers are: (1), (2), (4).