[分享]Some basic geometric theorems

[分享]Some basic geometric theorems

Raceleader 於 星期二 四月 08, 2003 7:33 pm


Congruent Triangle

SSS
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If AB=DE, BC=EF and CA=FD, then △ABC≡△DEF (SSS)


SAS
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If BA=ED, ∠BAC=∠EDF and AC=DF, then △ABC≡△DEF (SAS)


ASA
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If ∠ABC=∠DEF, BC=EF and ∠BCA=∠EFD, then △ABC≡△DEF (ASA)


AAS
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If ∠CAB=∠FDE, ∠ABC=∠DEF and BC=EF, then △ABC≡△DEF (AAS)


RHS
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If ∠BCA=∠EFD=90°, AB=DE and BC=EF, then △ABC≡△DEF (RHS)


Corresponding sides, congruent triangles, Corresponding angles, congruent triangles
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If △ABC≡△DEF:
Then AB=DE, BC=EF and CA=FD (Corresponding sides, congruent triangles)
Then ∠ABC=∠DEF, ∠BCA=∠EFD and ∠CAB=∠FDE (Corresponding angles, congruent triangles)

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Raceleader 於 星期二 四月 08, 2003 7:34 pm


Similar Triangle

AAA
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If ∠ABC=∠DEF, ∠ABC=∠DEF and ∠ABC=∠DEF, then △ABC∼△DEF (AAA)


3 sides proportional
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If AB/DE=BC/EF=CA/FD, then △ABC∼△DEF (3 sides proportional)


Ratio of 2 sides, included angle
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If AB/DE=CA/FD and ∠CAB=∠FDE, then △ABC∼△DEF (Ratio of 2 sides, included angle)


Corresponding sides, similar triangles, Corresponding angles, similar triangles
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If △ABC∼△DEF:
Then AB/DE=BC/EF=CA/FD (Corresponding sides, similar triangles)
Then ∠ABC=∠DEF, ∠BCA=∠EFD and ∠CAB=∠FDE (Corresponding angles, similar triangles)

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Raceleader 於 星期二 四月 08, 2003 7:35 pm


Adjacent angles on straight line
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If AOB is a straight line, then ∠AOC+∠COB=180° (Adjacent angles on straight line)

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Raceleader 於 星期二 四月 08, 2003 7:35 pm


Angles at a point
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If AO, BO, CO, DO and EO meet at O, then ∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360° (Angles at a point)

Proof
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Extend AO to F, so that AOF is a straight line.
AOF is a straight line (Given)
∠AOE+∠EOD+∠DOF=180° (Adjacent angles on straight line)
∠AOB+∠BOC+∠COF=180° (Adjacent angles on straight line)
∠AOB+∠BOC+∠COD+∠DOE+∠EOA=(∠AOE+∠EOD+∠DOF)+(∠AOB+∠BOC+∠COF)
∴∠AOB+∠BOC+∠COD+∠DOE+∠EOA=180°+180°=360°

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Raceleader 於 星期二 四月 08, 2003 7:36 pm


Vertical opposite angles
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If AB, CD meet at O, then ∠AOC=∠BOD and ∠AOD=∠BOC (Vertical opposite angles)

Proof
AOB and COD are straight lines (Given)
∠AOC+∠AOD=180° (Adjacent angles on straight line)
∠AOD+∠BOD=180° (Adjacent angles on straight line)
∠AOC+∠BOC=180° (Adjacent angles on straight line)
∴∠AOC=∠BOD and ∠AOD=∠BOC

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Raceleader 於 星期二 四月 08, 2003 8:02 pm


Parallel lines
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Definition: Two or more straight lines in the same plane that will never meet with each others

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Raceleader 於 星期二 四月 08, 2003 8:02 pm


Corresponding angles, AB//CD
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If AB//CD, then ∠EFB=∠FGD (Corresponding angles, AB//CD)

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Raceleader 於 星期二 四月 08, 2003 8:03 pm


Alternate angles, AB//CD
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If AB//CD, then ∠AFG=∠FGD (Alternate angles, AB//CD)

Proof
AB//CD (Given)
∠AFG=∠EFB (Vertical opposite angles)
∠EFB=∠FGD (Corresponding angles, AB//CD)
∴∠AFG=∠FGD

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Raceleader 於 星期二 四月 08, 2003 8:03 pm


Interior angles, AB//CD
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If AB//CD, then ∠BFG+∠FGD=180° (Interior angles, AB//CD)

Proof
AB//CD (Given)
∠EFB+∠BFG=180° (Adjacent angles on straight line)
∠EFB=∠FGD (Corresponding angles, AB//CD)
∴∠BFG+∠FGD=180°

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Raceleader 於 星期二 四月 08, 2003 8:03 pm


Corresponding angles equal
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If ∠EFB=∠FGD, then AB//CD (Corresponding angles equal)

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Raceleader 於 星期二 四月 08, 2003 8:04 pm


Alternate angles equal
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If ∠AFG=∠FGD, then AB//CD (Alternate angles equal)

Proof
∠AFG=∠FGD (Given)
∠AFG=∠EFB (Vertical opposite angles)
∴∠EFB=∠FGD
∴AB//CD (Corresponding angles equal)

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Raceleader 於 星期二 四月 08, 2003 8:04 pm


Interior angles supplementary
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If ∠BFG+∠FGD=180°, then AB//CD (Interior angles supplementary)

Proof
∠BFG+∠FGD=180° (Given)
∠EFB+∠BFG=180° (Adjacant angles on straight line)
∴∠EFB=∠FGD
∴AB//CD (Corresponding angles equal)

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Raceleader 於 星期二 四月 08, 2003 10:13 pm


Angle sum of triangle
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If ABC is a triangle, then ∠ABC+∠BCA+∠CAB=180° (Angle sum of triangle)

Proof
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Draw a line segment DE passes through A, such that DAE//BC.
DE//BC (Given)
∠DAB=∠ABC (Alternate angles, DE//BC)
∠EAC=∠BCA (Alternate angles, DE//BC)
∠DAB+∠CAB+∠EAC=180° (Adjacent angles on straight line)
∴∠ABC+∠BCA+∠CAB=180°

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Raceleader 於 星期二 四月 08, 2003 11:14 pm


Exterior angle of triangle
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ABC is a triangle.  If extend BC to D, then ∠ACD=∠ABC+∠BAC (Exterior angle of triangle)

Proof
∠ACD+∠BCA=180° (Adjacent angles on straight line)
∠ABC+∠BAC+∠BCA=180° (Angle sum of triangle)
∴∠ACD=∠ABC+∠BAC

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Raceleader 於 星期二 四月 08, 2003 11:36 pm


Convex polygon
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Definition: All interior angles of the polygon are less than 180°

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Raceleader 於 星期三 四月 09, 2003 12:13 am


Angles sum of polygon
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The sum of interior angles of a n-sided convex polygon is (n-2)180° (Angles sum of polygon)

Proof
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We can draw all diagonals of the n-sided convex polygon from a vertex.
We then find that (n-2) triangles are divided.
Each angle sum of triangle is 180° (Angle sum of triangle)
∴The sum of interior angles of a n-sided convex polygon is (n-2)180°

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Raceleader 於 星期三 四月 09, 2003 12:23 am


Sum of exterior angles of polygon
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The sum of exterior angles of a n-sided convex polygon is 360° (Sum of exterior angles of polygon)

Proof
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The angle sum of a n-sided convex polygon=(n-2)180° (Angles sum of polygon)
For every straight angle:
Interior angle+Exterior angle=180° (Adjacant angles on straight line)
∵There are n exterior angles (Given)
∴There are n straight angles
∴The sum of exterior angles of a n-sided convex polygon=n straight angles-angle sum of a n-sided convex polygon
∴The sum of exterior angles of a n-sided convex polygon=180°n-(n-2)180°=360°

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Raceleader 於 星期三 四月 09, 2003 12:42 am


Right-angled triangle
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Definition - One angle in the triangle is right angle

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Raceleader 於 星期三 四月 09, 2003 12:48 am


Pythagoras Theorem
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If ABC is a right-angled triangle, and ∠ABC=90°, then AB2+BC2=CA2 (Pythagoras Theorem)

Proof
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Draw a right-angled triangle CDE, such that △ABC≡△CDE, and BCD forms a straight line.  Join  
AE.
Let AB=c, BC=a, CA=b.
△ABC≡△CDE (Given)
∴AB=CD=c, BC=DE=a and CA=EC=b (Corresponding sides, similar triangles)
∴∠CAB=∠ECD, ∠ABC=∠CDE=90° and ∠BCA=∠DEC (Corresponding angles, similar triangles)
∠BCA+∠CAB=180°-∠ABC=180°-90°=90° (Angle sum of triangle)
∠BCA+∠ECD=180°-∠ACE (Adjacent angles on straight line)
∴∠ACE=90°

Area of trapezium ABDE=(1/2)(AB+DE)(BD)=(1/2)(c+a)(a+c)=(1/2)(a+c)2
Area of △ABC=Area of △CDE=(1/2)(AB)(BC)=(1/2)ac
Area of △ACE=(1/2)(AC)(CE)=(1/2)b2
∵Area of trapezium ABDE=Area of △ABC+Area of △CDE+Area of △ACE
∴(1/2)(a+c)2=(1/2)ac+(1/2)ac+(1/2)b2
∴a2+2ac+c2=2ac+b2
∴a2+c2=b2
∴AB2+BC2=CA2

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Raceleader 於 星期三 四月 09, 2003 1:14 am


Converse of Pythagoras Theorem
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If AB2+BC2=CA2, then ∠ABC=90° (Converse of Pythagoras Theorem)

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