Congruent triangles proved the problem.

1. Proof of the relationship between the sides of a triangle

Example 1

As shown in the figure below, ∠ ACB = ∠ ADB = 90, AC=AD, and E is any point on AB. Proof: CE=DE.

analyse

We carefully observe the figures and conditions, and by proving that two triangles are identical, we think that two straight lines are equal. So which two triangles are congruent? Since AC=AD and AE are common * * * edges, we only need to know ∠ 1 = ∠ 2. Since there is also the condition that ∠ ACB = ∠ ADB = 90 in the known conditions, we immediately thought of proving ∠ 60 by proving the congruence of two right-angled triangles.

Prove:

In Rt△ABC and Rt△ABD, AB=AB, AC=AD,

∴

Rt△ABC≌Rt△ABD(HL)。

∴

∠ 1=∠2.

In △ACE and △ADE,

∴

△ACE?△ADE(SAS)。

∴

CE=DE。

Example 2 as shown in the figure, in △ABC, extend the midline BE to G on AC side, make EG=BE, extend the midline CD to F on AB side, make DF=CD, and connect AF and AG.

(1) Complete the graphics as required and mark the letters;

(2) 2) What is the relationship between AF and AG? Prove your conclusion.

analyse

It's easy to draw a picture according to the meaning of the question. Then point D is the midpoint between AB and CF, and point E is the midpoint between BG and AC. We can easily judge △ ADF △ BDC and △ age △ CBE, so we can see that the relationship between AF and AG is equal.

Solution: (1) Complete the graph, as shown in the figure.

(2) The relationship between AF and AG is AF=AG.

It is proved that in △ADF and △BDC,

∴

△ADF?△BDC(SAS)，

∴

AF=BC。

It can also be proved that:

△ age△ ?△CBE(SAS).

∴

AG=BC，

∴

AF=AG。

2. Proof of angle relation in triangle

Example 3

As shown in the figure, it is known that AD bisects ∠BAC, EF bisects AD vertically, the extension line of intersection BC is at F, and AF is connected. Verification: ∠B=∠CAF.

analyse

If you divide AD vertically by EF, you can find ∠ ADF = ∠ FAD, and there are ∠CAF, ∠CAF=∠DAF-∠DAC in the conclusion to be proved, so it is easy to prove the conclusion here.

Evidence:

EF bisects AD vertically,

∴

FA=FD，

∴

∠FAD=∠ADF (equilateral and equiangular).

∵

∠B=∠ADF-∠BAD，∠CAF=∠FAD-∠DAC，∠BAD=∠DAC，

∴

∠B=∠CAF。

Example 4

As shown in the figure, in equilateral △ABC, D is the moving point on the side of AB, CD is one side, equilateral △EDC is upward, and AE is connected. Prove: AE‖BC.

analyse

In order to prove AE‖BC, we obviously want to find equal isosceles angle, internal dislocation angle or complementary ipsilateral internal angle. Through observation, we naturally want to find the internal dislocation angle equal to ∠EAC and ∠ACB.

Evidence:

△ABC and△△ EDC are equilateral triangles,

∴

∠ Equivalent cyclic density =∠ACB=60.

∵

∠ECD-∠ACD=∠ACB-∠ACD, that is ∠ACE=∠BCD.

It's also VIII

AC=BC，EC=DC，

∴

△ACE?△BCD。

∴∠EAC=∠B=60。

∴∠EAC=∠ACB.

∴

BC.

Example 5

As shown in the figure, in Rt△ABC, AB=AC, ∠ BAC = 90, and O is the midpoint of BC. If point M and point N move on lines AB and AC respectively and keep AN=BM when moving, please judge.

Break the shape of △OMN and prove your conclusion.

analyse

Since AN=BM, we will think that triANgles with line segments An, BM, ON and ∠NOM have congruence relations, so we think that it is easy to get △ nao △ MBO by connecting AO. We can conclude that △OMN is an isosceles triangle. When we think about this step, we should further consider whether it is an equilateral triangle or an isosceles right triangle and the angle obtained by △ nao △ MBO.

Solution: △OMN is an isosceles right triangle. The reason for this is the following：

∵

AB=AC，∠BAC=90，

∴

∠B=∠C=45。

∵

O is the midpoint of BC,

∴

∠NAO=∠OAB=∠CAB=×90 =

45，∠AOB=90。

∴

∠OAB =∠ Oba