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數 Deductive Geometry 聯

Contents 錄 : Deductive Geometry Summary of Geometry Proving Skills Method for solving Geometry Questions Construction of Geometry figures Deductive Geometry 1

(A) Summary of Geometry Proving Skills 1. Relationship between parallel lines and its angles. 行. i. If a = b,then AB // CD. (Corr. s equal.) a = b, AB // CD. (.) ii. If b = d,then AB // CD. (Alt. s equal.) b = d, AB // CD. (.) iii. If b + c = 180,then AB // CD. (Int. s supplementary) b + c = 180, AB // CD. (.) iv. If AB // CD, then a = b. (Corr. s, AB // CD, AB // CD ) a = b. (, AB // CD ) v. If AB // CD,then b = d. (Alt. s, AB // CD, AB // CD ) b = d. (, AB // CD ) vi. If AB // CD,then b + c = 180. (Int. s, AB // CD ) If AB // CD,then b + c = 180. (, AB // CD ) Deductive Geometry 2

2. Properties of angles at a point, on a straight line and in a triangle.,. ii. a + b + c + d = 360. (Angles at a point) ii. 180 a + b + c + d =. (Adjacent Angles on a straight line) iv. a = b. (Vert. Oppsite angles equal) vi. 180 a + b + c =. (Angles sum of triangle.) v. c = a + b (ext. angle of a triangle) vi. ABC is an isosceles AB = AC (Base sides of ). b = c (Bases s of ) vii. ABC is an equilateral AB = BC = AC (sides of are equal). a = b = c = ( s 60 of are equal). Deductive Geometry 3

3. Congruent Triangles. i. AB = DE and BC = EF and AC = DF ABC DEF (SSS) ii. AB = DE and BC = EF and ABC = DEF ABC DEF (SAS) iii. ABC = DEF and BC = EF and ACB = DFE ABC DEF (ASA) iv. ABC = DEF and ACB = DFE and AC = DF ABC DEF (AAS) v. ABC = DEF = 90 and AC = DF and BC = EF ABC DEF (RHS) 4. Similar Triangles. i. ABC = DEF and ACB = DFE and BAC = EDF ABC DEF (AAA) ii. AB DE AC BC = = ABC DEF (3 sides proportional) DF EF iii. AB AC = and BAC = EDF DE DF ABC DEF (ratio of 2 sides, included ) iv. ABC = DEF and BAC = EDF ABC DEF (A.A. Similar) Deductive Geometry 4

5. Theorem between the sides and angles in a triangle 度 理 i. AB + BC > AC (Triangle Inequality 不 ) ii. AB > BC ACB > BAC (Bigger interior, Longer corr. sides) iii. ACD > ABC and ACD > BAC (ext. > corresponding Interior ) iv. BD = DC AD is the median of ABC. v. BAD = CAD AD is the angle bisector of A in ABC. vi. AD BC AD is the altitude of A in ABC. 6. Bisectors i. BD is the angle bisector of ABC ABD = DBC ii. CD is the perpendicular bisector of AB. AC = BC and ACD =. 90 Deductive Geometry 5

7. Point of Intersection of straight lines i. I is the point of intersection of AB and CD. ii. AB, CD and EF are concurrent at point I. iii. AB, CD and EF are not concurrent. iv. O is the In centre of ABC AO, BO and CO are the angle bisector of A, B and C respectively. v. O is the Circumcenter of ABC DO, EO and FO are the perpendicular bisector of AB, BC and AC respectively. vi. O is the Centroid of ABC DO, EO and FO are the perpendicular bisector of AB, BC and AC respectively. vii. O is the Orthocenter of ABC DO, EO and FO are altitude of ABC. Deductive Geometry 6

(B) Method for solving Geometry questions 1. There are mainly two methods for solving Geometry questions. 兩. i. Bottom-up Approach ( ). ii. Association Approach ( 聯 ). iii. Applying the answers from the previous parts ( ). 黎 Sir : 1. Bottom-up Approach ( ): 2. Association Approach ( 聯 ): 3. Applying the answers from the previous parts ( ): 4. 羅, : 5. Deductive Geometry( ) Deductive Geometry 7

e.g. 1 例 2 In the figure above, prove that AD 2 = BC CE DC (5 marks) 2, AD 2 = BC CE DC (5 ) 2 2 To prove : AD = BC CE DC AC CE = ( ) BC AC 度, 2 AD = BC CE DC 2 Prove ( ) 藍 : Bottom-up Approach Solutions : : Association Approach 聯. AD 2 BC CE DC 2 = ------------------ (1A) 1. (side ) 2 + 2. Ratio of sides 例 Deductive Geometry 8

e.g. 2 例 After teaching 10 lessons, Andy is very hungry and so he eats a sandwich. When he eat a part of the sandwich (as shown in the figure), he find that the sandwich goes bad. So he throws it away. Find out the area of the part of sandwich where Andy ate. (8 marks) 10, Andy 了. 了, 來 了. 便. Andy 了. (8 ) : ( ) To find : Area of CDE = and = and = 度, Area of the CDE Prove = and = 藍 : Bottom-up Approach : Association Approach 聯. Deductive Geometry 9

Solutions : 1., : 2., 兩 + Deductive Geometry 10

e.g. 3 例 In the figure above, prove that ADC = BCD. (6 marks), ADC = BCD. (6 ) To prove : 度, ADC = BCD Prove 藍 : Bottom-up Approach Solutions : : Association Approach 聯. 1., 兩 + Deductive Geometry 11

e.g. 4 例 Find x. x. (3 marks / 3 ) 度, x =? Adding Find : x 藍 : Bottom-up Approach : Association Approach 聯. Solutions : 1., Deductive Geometry 12

e.g. 6 例 六 In the figure, ABC, AOE and BOD are straight lines and x y = 50. Find r + s, ABC, AOE BOD = 50 x y. s. (3 marks) r +. (3 ) Find : r + s 度, r + s =? r, s r, s 藍 : Bottom-up Approach : Association Approach 聯. Solutions : 1. (ext. ), 2., Deductive Geometry 13

e.g.7 例, AD BC, AG GC, x DF, CG//DF, BDEG. AB=8, BC=2. x. (8 ) In the figure, AD BC, AG GC, x DF, CG//DF, BDEG is a square. AB=8m BC=2. Find x. (8 marks) Find : x,,,,,, 度, x =? Solving! 藍 : Bottom-up Approach : Association Approach 聯. Deductive Geometry 14

Solutions : 1. 行 2. 行 3. 4. 3 equations, 3 unknowns 數 5. Deductive Geometry 15

e.g.8 例 In the figure, CD AB, ACB = 90, AD = 5cm, DB = 3cm, find CD., CD AB, ACB = 90, AD = 5cm, DB = 3cm, Find : CD. :, ( ) CD, AD DB 度, CD =? Prove 藍 : Bottom-up Approach : Association Approach 聯. Solutions : Consider ACD and BAD 1., 兩 + Deductive Geometry 16

e.g. 9 例 In the figure, AB=AC, AD=AE, Prove BF=CF., AB=AC, AD=AE, BF=CF. (3 marks) (3 ) Prove :, : ( ) BF CF : ( ) DBF, AB ECF, AC 度, BF = CF Prove 藍 : Bottom-up Approach : Association Approach 聯. Deductive Geometry 17

Solutions : 1., 兩 + Deductive Geometry 18

e.g. 10 例 In the figure, AB=AC, ABE = ACD. Prove that AD = AE. (2 marks), AB=AC, ABE = ACD. AD = AE. (2 ) : ( ) AD AE Prove : 度, AD = AE Prove, 藍 : Bottom-up Approach : Association Approach 聯. Solutions : 1., 兩 + Deductive Geometry 19

e.g. 11 例 In the figure, AB=BC. Prove BC // DE., AB=BC. BC // DE. (4 marks) (4 ) : ( ) Prove : BC // DE EFB CBF 度, BC // DE, 藍 : Bottom-up Approach : Association Approach 聯. Solutions : 1. 行 Deductive Geometry 20

e.g. 12 例 In the figure, O is the center of the circle. Prove ACB = 90 (4 marks), O. ACB = 90. (4 ) Prove : ACB = 90, 度, ACB = 90 藍 : Bottom-up Approach : Association Approach 聯. Deductive Geometry 21

Solutions : 1. 2. Deductive Geometry 22

(C) Construction of Geometry 1. Construction of Angle bisector of an angle. Step 1 Take B as the center. Choose an appropriate radius, and draw an arc to intersect BA and BC at P and Q respectively. Step 2 Take P and Q as centers, select a radius longer than 2 1 PQ and draw two arcs such that these two arcs intersect at D. Step 3 Join BD. BD is the angle bisector of ABC Deductive Geometry 23

2. Construction of perpendicular bisector of a line segment. Step 1 Take A as the center. Choose a 1 radius longer than AB and 2 draw an arc above and below the line segment AB. Step 2 Take B as the center. Use the same radius and draw another two ones to intersect the arcs at C and D. Step 3 Join CD. CD intersects AB at M, M is the mid-point of AB. CD is the perpendicular bisector of AB Deductive Geometry 24

3. Construction of a line passing through a given point on a line segment and perpendicular to that line segment. Step 1 Take P as the center. Choose an appropriate radius and draw an arc to intersect AB at H and K. Step 2 Take H and K as centers. Use a 1 radius longer than HK and 2 draw two arcs such that they meet at Q. Step 3 Join PQ. Then PQ AB. Deductive Geometry 25

4. Construction of a line passing through a point lying outside a line segment and perpendicular to that line segment. Step 1 Take P as the center. Choose an appropriate radius and draw an arc to intersect AB at H and K. Step 2 Take H and K as centers. Choose a radius longer than 1 HK and draw two arcs such 2 that the two arcs meet at Q. Step 3 Join PQ. The line PQ intersects AB at M, and PQ AB Deductive Geometry 26

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