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Mathematics: solving problems in plane geometry
plane geometry

Four important theorems:

Menelaus Theorem (Mei's Line)

△ If there are points P, Q and R on the three sides BC, CA and AB of△ △ABC or its extension line, the necessary and sufficient conditions for straight lines P, Q and R*** are as follows.

Ceva Theorem (Ceva Point)

△ABC has points P, Q and R on three sides of BC, CA and AB, then the necessary and sufficient conditions for points AP, BQ and CR*** are as follows.

ptolemy's theorem

The necessary and sufficient condition for the sum of the products of two opposite sides of a quadrilateral to be equal to its diagonal product is that the quadrilateral is inscribed with a circle.

Siemsen Theorem (siemsen Line)

The necessary and sufficient condition for drawing a perpendicular from a point to three sides of a triangle is that the point falls on the circumscribed circle of the triangle.

Example:

1. Let AD be the center line of the side BC of △ABC, and the straight line CF intersects with AD at F, which proves that:

Analysis of CEF section △ABD→ (Meyer theorem)

The commentary can also add auxiliary lines to prove that one of A, B and D is a parallel line of CF.

2. The straight lines passing through the gravity center G of △ABC intersect with CB in AB, AC, F and D in E respectively.

Prove:

Analyze the link and extend the intersection of AG and BC to m, then m is the midpoint of BC.

DEG section △ABM→ (Meyer theorem)

DGF section △ACM→ (Meyer theorem)

∴=== 1

Comment on Meyer theorem

3.d, E and F are respectively beside BC, CA and AB of △ABC.

, AD, BE and CF cross into △LMN.

Find LMN.

analyse

Comment on Meyer theorem

4. Based on the sides of △ABC, make similar isosceles △BCE, △CAF and △ABG outward. Prove that AE, BF and CG intersect at one point.

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Comments on Seva Theorem

5. It is known that ∠B=2∠C in △ABC. Verification: AC2 = AB2+ AB BC.

It is analyzed that parallel lines with A as BC intersect △ABC, and the circumscribed circle is D, which connects BD. Then CD=DA=AB, AC=BD.

According to Ptolemy theorem, AC BD = AD BC+CD AB.

Comment on Ptolemy Theorem

6. The regular heptagon A1A2A3A4A5A7 is known.

Prove: (The 2nd1All-Soviet Mathematics Competition)

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Comment on Ptolemy Theorem

7. The extension line of the BC side height AD of 7.△ ABC circumscribes P, PE⊥AB is in E, and the extension line of ED is in F. ..

Verification: BC ef = BF ce+be cf.

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Comments on siemsen's Theorem (siemsen Line)

8. Diagonal AC and CE of regular hexagon ABCDEF are divided into AM: AC = CN: Ce = K by straight lines of inner points M and N, and B, M and N*** respectively. Find K. (23-IMO-5)

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Comment area method

9.O is a point within △ABC, the distances from O to BC, CA and AB are represented by da, db and dc respectively, and the distances from O to A, B and C are represented by Ra, Rb and Rc.

Verification: (1) a ra ≥ b db+c dc;

(2)a Ra≥c d b+ b DC;

(3) Ra+Rb+Rc≥2(da+db+dc).

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Comment area method

At 10. △ ABC, H, G and O are vertical center, center of gravity and outer center respectively.

Verification: H, G, O three-point * * * line, HG=2GO. (Euler line)

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Comment on the same law

In 1 1. △ ABC, AB=AC, AD⊥BC = D, BM, BN ∠ABC, intersecting with AD in M and N, and extending CM and AB in E.

Verification: MB//NE.

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Comment symmetry transformation

12.g is the center of gravity of △ABC. With AG as the chord, BG is tangent to G, and the intersection of CG extends to D ... Verification: AG2 = GC GD.

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Comment translation transformation

13.c is the point above ⊙O with diameter AB=2, and p is within △ABC. If the minimum value of PA+PB+PC is 0, find the area s of △ABC at this time.

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Annotation rotation transformation

Fermat point: O is a point within △ABC, ∠ AOB = ∠ BOC = ∠ COA =120; P is any point in △ABC. Proof: PA+PB+PC≥OA+OB+OC. (O is fermat point)

Analysis will cc', OO', PP', link OO', PP'. Then △B OO' and △B PP' are regular triangles.

∴OO'=OB,PP' =PB. Obviously △ bo' c' △ BOC, △ BP' c' △ BPC.

Because ∠ bo' c' = ∠ BOC =120 =180-∠ bo' o, ∴A, o, o', c' four-point * * line.

∴ AP+PP'+p' c' ≥ AC' = ao+oo'+o' c', that is, PA+PB+PC≥OA+OB+OC.

14. (National Competition' 95) The inscribed circle O of rhombic ABCD intersects with all sides at E, F, G and H respectively, and the tangent of ⊙O on arc EF and arc GH intersects with AB, BC, CD and DA at M, N, P and Q respectively.

Authentication: MQ//NP.

Analysis from AB‖CD: To prove MQ‖NP, you only need to prove ∠AMQ=∠CPN.

Combining ∠A=∠C, you only need to prove it.

△AMQ∽△CPN

←,AM CN=AQ CP .

Connect AC and BD, the intersection point is the center o OF inscribed circle, let MN and ⊙O cut at k, and connect OE, OM, OK, on and of. Remember ∠ABO=φ, ∠MOK=α, ∠KON=β, then

∠EOM=α,∠FON=β,∠EOF=2α+2β= 180 -2φ.

∴∠BON=90 -∠NOF-∠COF=90 -β-φ=α

∴∠cno=∠nbo+∠nob=φ+α=∠aoe+∠moe=∠aom

And ∠ocn =∠ Mao, so,

∴AM CN=AO company

In the same way, AQ CP = ao company

To annotate ...

15. (National Competition' 96) ⊙O 1 and ⊙O2 are all tangent to the straight lines on the three sides of Δ δABC, with E, F, G and H as tangents, and the extension lines of EG and FH intersect with P. Proof: PA⊥BC.

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To annotate ...

16.(99 National Games) As shown in the figure, in the quadrilateral ABCD, the diagonal AC bisector ∠BAD. Take a little e on the CD, BE and AC intersect at f, and extend DF to BC and G. Verify: ≈GAC =≈EAC.

Prove: by linking BD with AC and H, and using Seva theorem for △BCD, we can get

Because AH is the angle bisector of ∠BAD, from the angle bisector theorem,

Therefore, it is available.

The parallel line of AB passing through C intersects the extension line of AG at I, and the parallel line of AD passing through C intersects the extension line of AE at J.

Then,

Therefore, CI=CJ.

And because CI//AB, CJ//AD, ∠ACI=π-∠BAC=π-∠DAC=∠ACJ.

Therefore, △ ACI △ ACJ, thus ∠IAC=∠JAC, that is ∠GAC=∠EAC.

It is known that AB=AD, BC=DC, AC and BD intersect at O, and any two straight lines EF and GH passing through O intersect with four sides of quadrilateral ABCD at E, F, G and H. Add GF and EH, and pay BD to m and n respectively. Verification: OM = on. (5th CMO)

Proof: If △ E'H' △ E' OH', you only need to prove e', m, H' * * *, that is, the points of E'H, BO and GF.

Remember ∠BOG=α, ∠GOE'=β. To connect e' f to BO to K, just prove = 1(Ceva inverse theorem).

=== 1

Note: Zheng shape: a quadrilateral with one diagonal vertically bisecting the other diagonal.

Corresponding to the three lines of 2: ∠ e 'ob = ∠ FOB, e 'h, GF and BO in the 99 League, they are * * *. Proof: ∠ gob = ∠ h 'ob.

In fact, the above conditions are all necessary and sufficient conditions, and the conclusion is still valid when m is on the OB extension line.

Proof method: the same method.

Butterfly theorem: P is the midpoint of the chord AB of ⊙O, and two chords CD and EF of ⊙O are introduced through point P to connect DE and AB in M and CF and AB in N. Prove: MP=NP.

Let GH be the diameter of p, FF'F, obviously' ∑⊙o. p∈GH, ∴ PF' = PF. ∵ PFPF', PAPB, ∴∠ FPN = ∠ F 'PM, PF = PF'.

And FF'⊥GH, AN⊥GH, ∴ff' ab. ∴∠F'PM+∠MDF'=∠FPN+∠EDF'

= ∠ eff'+∠ EDF' = 180, ∴P, m, d, f' four-point * * * circle. ∴∠PF'M=∠PDE=∠PFN。

∴△PFN≌△PF'M,PN=PM。

The general conclusion of the commentary is: given a point P on the chord AB in a ⊙O with radius R, two intersecting chords CD and EF pass through P, even CF and ED intersect AB at M and N, given OP=r and the distance from P to the midpoint of AB is A, then. (Analytic proof: using the knowledge of conical system)