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Select the proof required then click mouse key to view proof. Menu. Theorem 1 Vertically opposite angles are equal in measure. Theorem 2 The measure of the three angles of a triangle sum to 1800 . Theorem 3 An exterior angle of a triangle equals the sum of the two interior opposite

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Select the proof required then click

mouse key to view proof.

Menu

Theorem 1 Vertically opposite angles are equal in measure.

Theorem 2 The measure of the three angles of a triangle sum to 1800 .

Theorem 3 An exterior angle of a triangle equals the sum of the two interior opposite

angles in measure.

Theorem 4 If two sides of a triangle are equal in measure, then the angles

opposite these sides are equal in measure.

Theorem 5 The opposite sides and opposite sides of a parallelogram are respectively equal in measure.

Theorem 6 A diagonal bisects the area of a parallelogram

Theorem 7 The measure of the angle at the centre of the circle is twice the

measure of the angle at the circumference standing on the same arc.

Theorem 8 A line through the centre of a circle perpendicular to a chord

bisects the chord.

Theorem 9 If two triangles are equiangular, the lengths of the corresponding sides are in proportion.

Theorem 10 In a right-angled triangle, the square of the length of the side opposite to the right angle

is equal to the sum of the squares of the other two sides.

Constructions

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90

90

45

45

135

135

0

0

180

180

Theorem 1: Vertically opposite angles are equal in measure

Use mouse clicks to see proof

1

4

2

3

To Prove:Ð1 = Ð3 and Ð2 = Ð4

Proof:Ð1 + Ð2 = 1800 …………..Straight line

Ð2 + Ð3 = 1800 ………….. Straight line

ÞÐ1 + Ð2 = Ð2 + Ð3

Þ Ð1 = Ð3

Similarly Ð2 = Ð4

Q.E.D.

Constructions

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Theorem 2 the measure of the three angles of a triangle sum to 180 0

4

5

3

1

2

Theorem 2: The measure of the three angles of a triangle sum to 1800 .

Use mouse clicks to see proof

Given: Triangle

Proof:Ð3 + Ð4 + Ð5 = 1800Straight line

Ð1 = Ð4 and Ð2 = Ð5 Alternate angles

ÞÐ3 + Ð1 + Ð2 = 1800

Ð1 + Ð2 + Ð3 = 1800

Q.E.D.

To Prove:Ð1 + Ð2 + Ð3 = 1800

Construction:Draw line through Ð3 parallel to the base

Constructions

Sketches

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90

45

135

3

0

180

1

2

4

Theorem 3: An exterior angle of a triangle equals the sum of the two interior

opposite angles in measure.

Use mouse clicks to see proof

To Prove:Ð1 = Ð3 + Ð4

Proof:Ð1 + Ð2 = 1800 …………..Straight line

Ð2 + Ð3 + Ð4 = 1800 ………….. Theorem 2.

Þ Ð1 + Ð2 = Ð2 + Ð3 + Ð4

Þ Ð1 = Ð3 + Ð4

Q.E.D.

Constructions

Sketches

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a

c

b

d

Theorem 4: If two sides of a triangle are equal in measure, then the angles

opposite these sides are equal in measure.

Use mouse clicks to see proof

4

3

Given:Triangle abc with |ab| = |ac|

To Prove:Ð1 = Ð2

2

1

Construction:Construct ad the bisector of Ðbac

Proof: In the triangle abd and the triangle adc

Ð3 = Ð4 …………..Construction

|ab| = |ac|………….. Given.

|ad| = |ad|………….. Common Side.

Þ The triangle abd is congruent to the triangle adc……….. SAS = SAS.

Þ Ð1 = Ð2

Q.E.D.

Constructions

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b

c

a

d

Theorem 5: The opposite sides and opposite sides of a parallelogram are respectively equal in measure.

Use mouse clicks to see proof

Given: Parallelogram abcd

To Prove:|ab| = |cd| and |ad| = |bc|

and Ðabc = Ðadc

3

4

Construction:Draw the diagonal |ac|

1

Proof: In the triangle abc and the triangle adc

2

Ð1 = Ð4 …….. Alternate angles

Ð2 = Ð3 ……… Alternate angles

|ac| = |ac| …… Common

Þ The triangle abc is congruent to the triangle adc……… ASA = ASA.

Þ |ab| = |cd| and |ad| = |bc|

and Ðabc = Ðadc

Q.E.D

Constructions

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Theorem 6 a diagonal bisects the area of a parallelogram

c

b

a

d

x

Theorem 6: A diagonal bisects the area of a parallelogram

Use mouse clicks to see proof

Given: Parallelogram abcd

To Prove:Area of the triangle abc = Area of the triangle adc

Construction:Draw perpendicular from b to ad

Proof: Area of triangle adc = ½ |ad| x |bx|

Area of triangle abc = ½ |bc| x |bx|

As |ad| = |bc| …… Theorem 5

Area of triangle adc = Area of triangle abc

Þ The diagonal ac bisects the area of the parallelogram

Q.E.D

Constructions

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a

o

r

c

b

Theorem 7: The measure of the angle at the centre of the circle is twice the

measure of the angle at the circumference standing on the same arc.

Use mouse clicks to see proof

To Prove:| Ðboc | = 2 | Ðbac |

5

2

Construction:Join a to o and extend to r

Proof: In the triangle aob

4

1

3

| oa| = | ob | …… Radii

Þ | Ð2 | = | Ð3 | …… Theorem 4

| Ð1 | = | Ð2 | + | Ð3 | …… Theorem 3

Þ | Ð1 | = | Ð2 | + | Ð2 |

Þ | Ð1 | = 2| Ð2 |

Similarly| Ð4 | = 2| Ð5 |

Q.E.D

Þ | Ðboc | = 2 | Ðbac |

Constructions

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L

a

o

r

90 o

b

Theorem 8: A line through the centre of a circle perpendicular to a chord

bisects the chord.

Use mouse clicks to see proof

Given: A circle with o as centre

and a line L perpendicular to ab.

To Prove:| ar | = | rb |

Construction:Join a to o and o to b

Proof: In the triangles aor and the triangle orb

Ðaro = Ðorb ………….90 o

|ao| = |ob|………….. Radii.

|or| = |or|………….. Common Side.

Þ The triangle aor is congruent to the triangle orb……… RSH = RSH.

Þ |ar| = |rb|

Q.E.D

Constructions

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Theorem 9 if two triangles are equiangular the lengths of the corresponding sides are in proportion

|ab|

|ab|

|ab|

|ac|

|ac|

|ac|

To Prove:

=

=

=

a

|bc|

|bc|

d

|de|

|ax|

|ay|

|df|

|de|

|df|

=

=

|ef|

|ef|

2

2

1

3

4

5

x

y

e

f

Þ

As xy is parallel to bc

1

3

Similarly

b

c

Theorem 9: If two triangles are equiangular, the lengths of the corresponding sides are in proportion.

Use mouse clicks to see proof

Given: Two Triangles with equal angles

Construction: On ab mark off ax equal in length to de.

On ac mark off ay equal in length to df

Proof:Ð1 = Ð4

Þ[xy] is parallel to [bc]

Q.E.D.

Constructions

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b

a

a

c

b

c

c

c

a

b

b

a

Theorem 10: In a right-angled triangle, the square of the length of the side opposite to the right angle is equal to the sum of the squares of the other two sides.

Use mouse clicks to see proof

Given: Triangle abc

To Prove:a2 + b2 = c2

Construction: Three right angled triangles as shown

Proof: Area of large sq. = area of small sq. + 4(area D) (a + b)2 = c2 + 4(½ab)

a2 + 2ab +b2 = c2 + 2ab

a2 + b2 = c2

Q.E.D.

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