The main content of this section is the nature of similar triangles, which is also one of the main contents of this chapter. On the basis of studying similar triangles's judgment, we will further study the nature of the triangle, so as to complete a comprehensive study of the definition, judgment and nature of similar triangles.
1. Properties of similar triangles
The angles of (1) similar triangles are equal, and the corresponding proportions are in direct proportion.
(2) The ratio of similar triangles to height, to centerline and to angular bisector are all equal to similarity ratio.
(3) The perimeter ratio of similar triangles is equal to the similarity ratio.
The above items can be summarized as follows: The ratio of corresponding line segments in similar triangles is equal to the similarity ratio.
(4) The ratio of similar triangles area is equal to the square of similar triangles similarity ratio.
2. The application of similar triangles property.
(1) can be used to prove that the line segments are proportional (or equal product line segments) and the angles are equal.
(2) Find out the unknown elements (edge, height, angle bisector, midline and angle) from some known elements in similar triangles.
(3) Used to calculate perimeter, area, etc.
(4) It is used to prove the equal division (or area ratio) of line segments.
Analysis of key and difficult points
Example 1 is shown in Figure 5.5- 1. It is known that △ ABC ∽△ A ′ B ′ C ′, points D and D ′ are the midpoint between BC and B ′ C ′, AE⊥BC is in E, and A ′ E ′⊥ B ′ C ′ is in E.
The analysis requires that △ADE is similar to △ a ′ d ′ e ′, and these two triangles are right triangles. To judge the similarity theorem of right triangle, it is only necessary to prove that the hypotenuse is proportional to the right triangle, and the hypotenuse and right triangle are just the median line and height of △ABC and △ A ′ B ′ C ′ (that is, the line segments corresponding to two similar triangles).
It is proved that △ ABC ∽△ a ′ b ′ c ′ ad and a ′ d ′ are the midline respectively, and AE and a ′ e ′ are high respectively.
∴= =∴rt△ade∽rt△a′d′e′
Example 2 is shown in Figure 5.5-2. At △ABC, EF‖BC, ef = BC = 2cm, the circumference of △AEF is 10cm. Find the perimeter of trapezoidal BCFE.
In the analysis, we can get = from EF= BC, that is, the similarity ratio, and then get the perimeter of △ABC from similar triangles property. The difference between the two perimeters plus the length of EF is the perimeter of the trapezoidal BCFE.
Solution: ef = BC ∴ =
∴△AEF∽△ABC BC
∴ = =
∴ =
∴△ABC circumference = 15 (cm)
∴ perimeter of trapezoidal BCFE = △ ABC perimeter -△AEF perimeter +2EF
= 15- 10+4=9 (cm)
Example 3 is shown in Figure 5.5-3. In △ABC, DE‖BC, s △ ade ∶ s △ ABC = 4 ∶ 9, ① find AE ∶ EC; ② Find S△ADE∶S△CDE.
In this paper, the properties of similar triangles, their combination ratio and the calculation formula of triangle area are analyzed. The ratio is obtained by =, and then the heights of AE∶EC, △ADE and △CDE are obtained from the proportional correlation properties, and AE∶EC is obtained from the triangle area calculation formula.
Solution: ① British Columbia ∴△ADE∽△ABC
∴ = ∴ =
∴ = =
Namely =
② Connect CD, cross D as DH⊥AC and cross AC as H.
= = =
Example 4 As shown in Figure 5.5-4, it is known that m is the midpoint of the AB side of □ABCD, and CM intersects BD at point E. What is the ratio of shadow area to parallelogram ABCD in the figure?
This is a comprehensive test, which examines similar triangles's properties, area calculation and equal product theorem. Let DN⊥AB be in n, e be in f and GF⊥AB be in f.
∫M is the midpoint of AB.
∴S△AMD=S△DMB= S△ABD= S□ABCD
∫s△MBD = s△MBC (two triangles with the same base height have the same area)
∴ s △ MBD-s △ MBE = s △ MBC-s △ MBE, that is, s △ DME = s △ CBE.
∵MB‖DC,∴△BEM∽△DEC
= =, therefore =
∵DN=GF,∴ =
Here we go again: = =
∴ =, that is, S△DME= S△MBD.
∴S△DME= × S□ABCD= S□ABCD
∴s△dme+s△bmc= s□ABCD+s□ABCD = s□ABCD
Therefore, the ratio of shadow area to parallelogram area in Figure 5.5-4 is.
Example 5 As shown in Figure 5.5-5, extend the side BC of the square ABCD to E, so that CE = AC, AE and DC intersect at point F, and find the value of CE∶FC.
This is an examination question to examine the ability to solve problems by using similar triangles's nature. Let the side length ABCD of a square be a, then AC = A, AB = A, Be =+1) a. 。
Solution: ∫DC‖ab, ∴△ECF∽△EBA, =, thus =+1,that is, EC ∶ FC = (+1):1.
Example 6 is shown in Figure 5.5-6, □ABCD, where E is a point above BC and AE intersects BD at point F. It is known that Be ∶ EC = 3 ∶ 1, S △ FBE = 18, and S△FDA is found.
By analyzing the given conditions, we can easily get △FBE∽△FDA. Then, from Be ∶ EC = 3 ∶ 1, we can easily get Be ∶ AD = 3 ∶ 4, because the area ratio of similar triangles is equal to the square of the similarity ratio, so we get S△FDA.
Solution: From □ABCD, get BE‖AD ∴△FBE∽△FDA.
Beryllium: cerium = 3:1Be: BC = 3: 4.
∵ BC = AD, ∴ ratio: AD = 3: 4.
∴ = () 2, which means = () 2.
∴S△FDA= =32。
A clever way to solve difficulties
Example 1 is shown in figure 5.5-7. In △ABC, ∠ ACB = 90, BC = 8 cm, AC = 6 cm, C is the center of the circle and CA is the radius, so what is the length of AD?
The analysis of this topic is a comprehensive question, and the knowledge points examined include similar triangles's judgment and nature, the nature of isosceles triangle, etc. For AD, we found that △CAD is an isosceles triangle after linking CD, and the length of the base of isosceles triangle is required. So, if we think of CE⊥AB and AE= AD in E, if we can find the length of AE, the problem will be solved. Therefore, we only need to prove △ AEC.
Solution: Make CE⊥AB in E key and connect CD.
CA = CD
∴ AE = AD, which means AD = 2AE.
AB = 10 is obtained from known conditions and Pythagorean theorem.
∠∠ACB =∠AEC，∠A=∠A
∴△AEC∽△ABC
∴ =
∴ ac2 = AE ab, that is, 62 = aex10.
So AE = 3.6 (cm)
∴AD=7.2(cm)
Example 2 is shown in Figure 5.5-8. In △ABC, DE‖BC, take a little f from AB to make S △ BFC = S △ Ade, and verify: Ad2 = AB BF.
Evidence: BC ∴△ ADC ∴△ ABC.
∴ =
∴s△ade=s△bfc :=
And = =
= = BF, that is, Ad2 = AB BF.
Hugging: The key to solve this problem is to find out the ratio formula or product formula by using the properties of similar triangles and the area calculation formula of triangle.
Example 3 is shown in Figure 5.5-9. Rectangular FGHN is inscribed with △ABC, f and g on BC, n and h on AB and AC respectively, AD⊥BC on D, NH on E, AD = 8 cm, BC = 24 cm, NF ∶ NH = 1 ∶ 2. Find the area of this rectangle.
Solution: ∴△ANH∽△ABC British Columbia
AE and AB are the heights of △ANH and △ABC respectively.
∴ =
Let nf = x, then NH = 2x.
AE=AD-ED=8-x
∴ =
Solution: x = 4.8
∴2X=9.6
∴S rectangular coordinate ABCD = NH NH = 9.6× 4.8 = 46.08 (cm) 2.
Pointing: A proportional formula is obtained by using similar triangles's properties, and then several unknowns of the proportional formula are transformed into an unknown number through quantity conversion, and some calculation problems are solved by algebraic methods. Is an important way to solve the problem.
Textbook problem solving
Example 1 is shown in Figure 5.5- 10. In rectangular ABCD, AB=a, BC = B, M is the midpoint of BC, and DE⊥AM and E are vertical feet. Verification: DE=. (P248B.2)。
Analysis shows that AD ∶ AM = DE ∶ AB can be obtained from △ADE∽△MAB, and DE is associated with A and B. 。
Proof: according to the right-angle ABCD, ∠ B = 90 ad ‖ BC.
∴∠DAE=∠AMB
∵DE⊥AM ∴∠DEA=∠B=90
∴△ade∽△mab :=
Ad = a, ab = b, and m is the midpoint of BC.
∴AM= = =
∴DE= =
Propositional trend analysis
In this part, the key point of the senior high school entrance examination is to comprehensively use similar triangles's judgment, property theorem and other geometric knowledge to calculate and prove, usually to prove proportional line segments and equal product line segments, and to find the side length and area of triangles.
Typical hot issues
Example 1 As shown in Figure 5.5- 1 1, in □ABCD, AE ∶ EB = 1 ∶ 2, and S △ AEF = 6cm2, then the value of S△CDF is ().
a . 12 cm2 b . 24 cm2 c . 54 cm2 d . 15 cm2
The analysis is similar to the above example, but there are some changes. AE ∶ EB = 1 ∶ 2, AE ∶ AB = 1 ∶ 3.
Solution: ∫□AB=CD, ab = CD, ∴ AE ∶ CD = 1 ∶ 3.
∵AE‖CD,∴△AEF∽△CDF
∴ =( )2,
That is = () 2.
∴ s △ CDF = 54 (cm) 2, so C.
Example 2 is shown in Figure 5.5- 12. Find the value of AC at △ABC, AB = 7, AD = 4, ∠ ACD = ∠ B.
Analyze this topic and examine the ability to apply the basic properties of similar triangles.
Solution: ∫∠A is the angle ∠ ACD = ∠ B,
∴△ACD∽△ABC ∴ =，
That is ac2 = ad ab.
∴AC=
=
= 2 (minus the root).
Example 3 is shown in fig. 5.5- 13, at △ABC, DE‖BC, S△ADE∶S quadrilateral BCED = 1 ∶ 2, BC = 2. Find the length of DE.
Analyze this topic, and investigate the ability of applying similar triangles's properties to solve practical calculation problems.
∵DE‖BC,∴△ADE∽△ABC.
The length of DE is needed, because the length of BC is known, so only the value of similarity ratio is needed. According to S△ADE∶S quadrilateral BCED = 1 ∶ 2, S △ ade ∶ S △ ABC = 1 ∶ 3. Comparing the relationship between area ratio and similarity ratio in similar triangles, it is not difficult to find. Solve problems smoothly.
Solution: ∫S△ADE∶S quadrilateral BCED = 1 ∶ 2.
∫S△ADE∶S△ABC = 1∶3
And \de \u BC,
∴△ADE∽△ABC.∴DE∶BC= 1∶
∫ BC =2 ∴DE= =2
Example 4: As shown in Figure 4.4- 14, the vertex C passing through △ABC is a straight line that intersects with AB side and central line AD at points F and E respectively, and the intersection point D is DM‖FC and AB at points M. 。
(1) if S△AEF∶S quadrilateral MDEF = 2 ∶ 3, find AE ∶ ed;
(2) verification: AE FB = 2af ed.
Analysis (1) This is a comprehensive ability test. If there are parallel conditions in the test, you will find similar triangles. The area ratio given in the test can be converted into the area ratio of similar triangles. With the area ratio, the similarity ratio can be obtained, and then it can be solved by transformation (1). (2) Prove that equal product formula can be transformed into equal ratio formula, which is actually parallel lines.
Solution: (1)∫S△AEF∶S quadrilateral MDEF = 2 ∶ 3.
∴S△AEF∶S△ADM=2∶5
∫DM‖cf ∴△aef∽△adm
∴ =
= = =
So AE ∶ ed = (+2) ∶ 3.
(2) Proof: ∫DM‖cf∴=
∴ =
∫d is the midpoint of BC ∴M is the midpoint of FB, that is, 2fm = FB.
∴ =, that is, AE FB = 2af ed.
Intensive exercises this week:
Synchronous outline exercise
Fill in the blanks
1. If the ratio of corresponding sides in similar triangles is 1∶3, their area ratio is.
2. Given the similarity ratio of two similar triangles, the ratio of their corresponding heights is.
3. As shown in Figure 5.5- 15, in △ABC and △BED, if = = and the difference between the perimeters of △ABC and △BED is 10cm, then the circumference of △ABC is cm.
Figure 5.5- 15 Figure 5.5- 16
4. If the similarity ratio of two similar triangles is 2∶3 and the sum of their areas is 13cm2, then their areas are respectively.
5. As shown in Figure 5.5- 16, it is known that C is a point on the AB line, and both △ACM and △BCN are equilateral triangles. If AC = 3, BC = 2, BM passes through CN to D, and the area ratio of △MCD to △BND is.
6. If the height ratio of two similar triangles is 4: 5, then their area ratio is 4: 5.
7. The area ratio of two similar triangles is 1: 9, so their corresponding height ratio is.
8. As shown in Figure 5.5- 17, when △ABC, DE‖BC, =, and s △ ABC = 8cm2, then s △ ade = cm2.
9. If the similarity ratio of two similar triangles is 2∶3, then their area ratio is.
10. If the ratio of corresponding sides of two similar triangles is 4∶5 and the sum of perimeters is 18cm, then the perimeters of these two triangles are cm and cm respectively.
Second, multiple choice questions
1. as shown in figure 5.5- 18, DE‖BC, and =, then the area ratio of △ADE to △ABC is s △ ade: s △ ABC = ().
A.2∶5 B.2∶3 C.4∶9 D.4∶25
2. As shown in Figure 5.5- 19, △ABC∽△ACD, and the similarity ratio is 2, then the area ratio S△BDC∶S△DAC is ().
a . 4∶ 1 b . 3∶ 1 c . 2∶ 1d . 1∶ 1
3. Given that the perimeters of two similar triangles are 8 and 6 respectively, their area ratio is ().
A.4∶3 B. 18∶9 C.2∶ D
4. When the area ratio of two similar triangles is 1: 2, the perimeter ratio is ().
A.b . 1∶c . 1∶4d . 4∶ 1
5. As shown in Figure 5.5-20, in Rt△ABC, ∠ACB is a right angle, and CD⊥AB is in D. The following formula is wrong ().
A.AC2 = AD AB B . BC2 = BD BA C . CD2 = AD DB D . AB2 = AC BC
6. In Rt△ABC, ∠ AC=b = 90, CD⊥AB, the vertical foot is D, let BC = A, AC=b, if AB = 16, CD = 6, A-B = ().
A.4 B. 8 C.8 D.4
7. Two triangles that meet the following conditions must be congruent ()
A similar and corresponding median ratio is equal to1B. Both sides are equal to one of the diagonals.
C. these three angles are equal. D. the heights of the two sides and the third side are equal.
8. In the square of ABCD, e is the midpoint of AB and BF⊥CE is in F, then the square of S△BFC∶S ABCD is equal to ().
1∶3 b . 1∶4 c . 1∶5d . 1∶8
9. As shown in Figure 5.5-2 1, divide the height AD of △ABC into three equal parts, and the bisectors are parallel lines with the bottom, so that the triangle is divided into three equal parts. Let the areas of these three parts be S 1, S2 and S3 respectively, then S 1: S2: S3 is equal to ().
a . 1∶2∶3 b . 2∶3∶4 c . 1∶3∶5d . 3∶5∶7
10. As shown in Figure 5.5-22, if △ABC, ∠ CBA = 90, and BD⊥AC is in D, then the error in the following relationship is ().
A.BD2 = AC B. BD2 = DC C.AB2=AC2-BC2 D.AB2=AC BC.
Third, answer questions.
1. as shown in figure 5.5-23, ∠ 1 = ∠ 2, ∠ B = ∠B=∠D, AB = DE = 5, BC = 4.
(1) Verification: △ ABC ∽△ ade; (2) Find the length of AD.
2. As shown in Figure 5.5-24, in Rt△ABC, ∠ c = 90, and in D, ⊥ AB, verify that Cd2 = ad db.
3. As shown in Figure 5.5-25, in □ABCD, BC = 2ce, and find: S△CEF∶S□ABCD.
4. As shown in Figure 5.5-26, it is known that ED⊥AB, AC⊥EB, D, C are vertical feet, G is a point above DE, AG⊥BG, vertical feet are G.ED, and AC meets the f test: dg2 = de df.
5. As shown in Figure 5.5-27, in the isosceles triangle AB=AC, AB = AC, AD⊥BC in D, CG‖AB and BG in E and F respectively, it is proved that BE2 = EF EG.
Quality optimization training
As shown in Figure 5.5-28, in △ABC, BC = 24, height AD = 12, two vertices E and F of rectangular EFGH are on BC, and the other two vertices G and H are on AC and AB respectively, and EF ∶ EH = 4 ∶ 3. Find the length of ef and eh.
Practical application of life
There is a parallel section on both sides of a river. There is a tree every 5 meters on this side of the river, and there is a telephone pole every 50 meters on the other side of the river. Looking across the river 25 meters away from the shore, you can see that two poles adjacent to each other on the other side are just covered by two trees here, separated by three trees. Please ask the width of this river.
Knowledge inquiry learning
As shown in Figure 5.5-29, when aiming, it is required that the notch on the scale of the gun should be in a straight line along the center A, that is, the aiming point C (above), so as to hit the target. The baseline AB of submachine guns is known to be 38.5cm long. If the shooting distance AC = 100 m, when the aiming point deviation BB' in the gap is 1mm, the deviation will be hit.
Reference answer
I. 1. 1:92. 3. 25 4.4 square centimeters and 9 square centimeters 5.9: 46. 16: 257
.1:38.29.4: 910.8cm,10cm.
2.1.d2.b3.b4.b5.d6.b7.a8.c9.c10.d.
Three. 1.① Omit ②
2. Certificate △ACD∽△CBD
3. 1∶ 12
4. First prove △ADF∽△EDB and then prove △AGD∽△GBD.
5. Even EC, certificate △ FEC △ CEG
Quality optimization training ef = 9.6 eh = 7.2
The river is 37.5 meters wide.
The influence deviation cc' of knowledge inquiry learning is about 26.0cm.