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Infringement complaint *High school math contest tutorial flat geometry*

Lecture 1 Note the addition of parallel lines

In the same plane, two lines that do not intersect are called parallel lines. Parallel lines are the most basic and very important figure of junior high school plane geometry. When proving certain plane geometric problems, if appropriate, you can add the appropriate according to the needs of the certificate. Parallel lines can make the proof smooth and concise. Adding parallel line questions, there are generally four cases as follows.

1 To change the position of the corner

As you know, two parallel straight lines are intercepted by the third straight line, with the same position angles, the internal misalignment angles, and the complementary internal angles. Using these properties, you can often change the position of some corners by adding parallel lines to meet Need for solution.

Example 1 Let P and Q be two points on the line segment BC, and BP = CQ, A is a moving point outside BC (as shown in Figure 1). When point A moves to ∠BAP = ∠CAQ, what triangle is △ ABC? Prove your conclusion. Answer: When point A moves to 到 BAP = ∠CAQ, △ ABC is an isosceles triangle.

Proof: As shown in Figure 1, the parallel lines passing through points P and B for AC and AQ respectively can get intersection D. Connect DA.

In △ DBP = ∠AQC, obviously

∠DBP = ∠AQC, ∠DPB = ∠C.

D

A

BP Figure 1

Q

From BP = CQ, we can know △ DBP≌ △ AQC. There are DP = AC, ∠BDP = ∠QAC. So, DA∥BP, ∠BAP = ∠BDP.

Then A, D, B, and P have four points in a circle, and the quadrilateral ADBP is an isosceles trapezoid. So AB = DP. So AB = AC.

Here, by making a parallel line, ACQAC is “flattened” to the position of 的 BDP. The four points A, D, B, and P are co-circular, so the proof is smooth. ∠BAF = ∠BCE. Prove: ∠EBA = ∠ADE. E prove: as shown in Figure 2, the points A, B are made parallel lines of ED and EC, and the intersection point P is connected to PE. PBA≌ △ ECD. PA = ED, PB = EC. =

Obviously, the quadrilateral PBCE and PADE are parallelograms. There are ∠BCE = ∠BPE, ∠APE = ∠ADE. From ∠BAF = ∠BCE, we can know that ∠BAF = ∠BPE.

There are P, B, A, and E at four points. So, ∠EBA = ∠APE. So, ∠EBA = ∠ADE.

P

B Figure 2

DC

F

Here, by adding parallel lines, the four corners of the known and the unknown are closely connected by the four points of P, B, A, and E. ∠APE becomes an equal medium for ∠EBA and ∠ADE, and the proof is very clever. . 2 Want to "send" line segments everywhere

By using the two equal distances between parallel lines and equal parallel line segments sandwiched between parallel lines, it is often possible to "send" certain line segments to appropriate positions by adding parallel lines to prove the problem.

Example 3 In △ ABC, BD and CE are angle bisectors, and P is any point on ED. Pass P as the vertical line of AC, AB, BC, and M, N, and Q are vertical feet. Verify: PM + PN = PQ.

A

Proof: As shown in Figure 3, the parallel line passing through point P as AB intersects BD at F, and the parallel line passing through point F as BC intersects PQ and AC NM respectively.

E

At K, G, and PG. D is parallel from BD to ABC. It can be seen that the distance between points F and AB and BC is equal. There is KQ = PN. Obviously,

EPPD

B

K

Q Figure 3

C

=

EFFD

=

CGGD

It can be seen that PG∥EC.

From CE to ∠ BCA, GP to ∠ FGA. PK = PM. Therefore, PM + PN = PK + KQ = PQ.

Here, by adding parallel lines, PQ is "opened" into two sections, and it is proved that PM = PK, then PM + PN = PQ. The method is very simple. 3 For the conversion of line segment ratio

Because "a straight line parallel to one side of the triangle cuts the other two sides, the corresponding corresponding line segments are proportional." In some problems, you can add benign lines to achieve a benign transformation of certain line segment ratios. This is often encountered in plane geometry problems To.

Example 4 Suppose M1 and M2 are points on the BC side of △ ABC and BM1 = CM2. Ren Zuo always submits AB, AC, AM1, AM2 to P, Q, N1, and N2. Test:

ABAP

＋

ACAQ

=

AM

1

AN1

＋

AMAN

twenty two

.

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