初二数学下册知识点总结.pdf

初二数学下册知识点总结.pdf

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1、-1 = / M =Q T 1=Q TB T 0 S = Q T . i 1 H q 5 = Q T 2 B 1 d 0. 21 T 1 , 2 i P . 3 Z Z y T Z i c T 3 | S B 1 p . 4=Q TEE5. 5=Q T1 vlZE 1 1 v l 2 = Q T “ M = Q | = 1 v l 3 s Y Z 1 v l . 6 Z Z $ “ T Z “ “ T Z . 7=Q T“EE5 1 2 3 s s | S s 8 Z E s T s 0 s s y T P s M T . 8s y T 9 y T . 9Ke=Q T 1 / H q = Q

2、T S K e = Q T $ 7 Z y y T T $ 7 Z c ? 7 y y T 2 K e = Q T $ 7 Z ? c l a s 3 y T Q 2 O c s 3 e = Q T H 1 $ 7 Z 5 s y s y T 4 = Q T 9 K T A K e = Q T . -2 10=Q Te5+ 1 A H q 5 2 c H q 5 3 ) H q 5 . =Q T+ = Q T K e = Q T T $ 7 Z M + = Q T S = Q T . 12=Q T = Q T F a h a a “ a Z a 7 Z B - V S = B M T p =

3、Q T a 2 = Q T B 1 5 = Q T a e = Q T ? i “ E H s s e L P E T . H+ A) Q 1 p Y a a1+ 1 H = H = 360 2 H . + V r T (1)A+BC+D=360 (2)123460 2H = 1 nH = (n-2)180 2 i H 36. + V r T 3 H y ABCD H + V r T (1)ABCD H BC (2) H AB=CD= (3) H DAB=C (4) H OOD (5)ABC H +A=180 -3 4. H . + V r T (1)ABCDBC H H (2)= H ABC

4、D H (3) 5. y ABCD (2)13 + V r T (1) 2ABCD =D90 (3) ACB 6. H ABCD . (1)2 (3) + V r T (1)ABCD H =90 H (2)ABCD=90 H (3) 7 + y ABCD + + V r T (1) 2ABCD + =A (3) + ACBD=CDB -4 8 + H H ABCD + . + V r T (1)ABCD H = H + (2)ABCDA H + (3) H ACBD H + 9Z y ABCD Z 1 2 3 + V r T (1) 2ABCD Z =A =90 (3)ABCD Z =ABD

5、10Z H ABCD Z . (3)ABCD = H Z + V r T (1)ABCD H =ABC=90 H Z (2)ABCD + =90 H Z 1%0+ V r T -5 y ABCD % 0 (1)ABCD % 0 AB=CD (2) % 0 BAD=C (3) % 0 12%0 H ABCD % 0 (3) 0 O BC AC=BD H % 0 + V r T (1)ABCD 0 O ADBC = H % 0 (2)ABCD 0 O ADBC = H % 0 13 LsL w T B F L B H L L M * L L 9 M 2 V 0 B % L A s 6 B % m

6、3 V B H 6 B H L A s H . m (2) (3) + V r T (1) 2ABCD 0 O ABCD E=F F (3)ADB EC = 14 L L H i O B . + V r T AD=BEC O =B 150L 0 L i O + V r T ABCD 0 O ABCD -6 B . DE=ACF=B F O E=(B+CD) + + B) Q) Q 1 p aa1A b45 1 p aa1A b45 B Q H H = H H L W H + Z m 0 % 0 0 L 0 L . = 1 1 1 m . 2 1 m L V i O $ s . 3 T m L

7、V B i O $ B s * m 1 B . T 1 S + =abch. aa b + L ,c + H h cH 2 H . H H h a 3 S0 = a+b h=L. aa b 0 0 ,L 0 L M 1 n H H 5 L H T . 2 ? 5 m | B “ B B M ”. 3 m H a a + a Z V 1 “ . 4 n m m a % a H a H a % 0 m H m L a a + a Z a H a . i L H . 5 0 n L -7 6 + n T 1 5 m ABCD H O E F * ABC=FD. m ABC B=90 O CD * C

8、=D. m ABCD + O E * =2. m ABC O BE D * AC=. m ABCD 0 Ea F % O GBC * EFAG= D+BC A. m . m ADBC * 1 S=S 2 . M M + + ) Q) Q 1 p Y a a1+ 1 p Y a a1+ -8 1“ 1 ” # I B H L H H L L 1 2 T B H L H H L L 1 * H L H . 1 3 2 + V r T (1)DEBC (2)DEBC (3) DEBC 21 1 1 a:b=cdad=bc 2 1 T * 3 1 T * . “ ”M B H L H H L M M

9、. + V r T DEBC A 4 “A”M + V r T -9 T B 6 B M * M . A= EDCB 5 “SA”M T B H H 6 B H H 1 i O C M * M . + V r T A= DEBC 6“ ”M # 1 $ | H s M 2 m H H | H | H 1 | H s | H H L 1 . + V r T (1)A CDB A (2)CBD A 2 = BCD 2 7M 1 M M H 1 2 M 1 L 1 s L a 1 M 1 3 M 1 M 1 Z . (1)ABCEFG BAC=FEG (2)ABCEFG Da H L (3)ABCE

10、FG + + ) Q) Q 1 p aa1A b45 1 p aa1A b45 B Q 1 L a 1 a 1 a s a M a M 1 . -10- = 1 L s L 1 H L H L L 1 . 2 “ ” 1 B H i O H M L H H 1 . 3 “S” M T B H H 6 B H H 1 * M . 4 “HL” M T B | H B H H 6 B | H B H H 1 * M . M 1 T L / M X / L L . 2 L 1 5 s Z E 1 E 1 p 1 T ? s B i s M V 7 P 1 T 2 L E 1 T ? s V m M L 1 T L B H + H 1 T s M 3 1 E W 1 E E W E E % 1 L 5 O 5 “ M V I n 1 E M H m M L W 1 1 H V O V 7 w 4 L s E M H 1 Z i p L n 5 “ 5 “ C 1 T V 5 “ X L p L ? s ? M V H 1 Z p L . 3 M . 1 2 3 13