1. Menu Theorem 1 Vertically opposite angles are equal in measure.
Theorem 2 The measure of the three angles of a triangle sum to 180o .
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 The opposite sides and opposite sides of a parallelogram are respectively equal in measure.
Theorem A diagonal bisects the area of a parallelogram
Theorem 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 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 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.

2. Theorem 1: Vertically opposite angles are equal in measure
180 90 45
135
180 90 45
135 1 4 2 3
To Prove: Ð1 = Ð3 and Ð2 = Ð4
Proof: Ð1 + Ð2 = ………….. Straight line
Ð2 + Ð3 = ………… Straight line
Þ Ð1 + Ð2 = Ð2 + Ð3
Þ Ð1 = Ð3
Similarly Ð2 = Ð4
Q.E.D.
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3. Theorem 2: The measure of the three angles of a triangle sum to 1800 .
Given: Triangle
To Prove: Ð1 + Ð2 + Ð3 = 1800 1 2 3
Construction: Draw line through Ð3 parallel to the base 4 5
Proof: Ð3 + Ð4 + Ð5 = 1800 Straight line
Ð1 = Ð4 and Ð2 = Ð5 Alternate angles
Þ Ð3 + Ð1 + Ð2 = 1800
Ð1 + Ð2 + Ð3 = 1800
Q.E.D.
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4. Theorem 3: An exterior angle of a triangle equals the sum of the two interior
opposite angles in measure. 1 2 3 4
180 90 45
135
To Prove: Ð1 = Ð3 + Ð4
Proof: Ð1 + Ð2 = ………….. Straight line
Ð2 + Ð3 + Ð4 = ………….. Triangle.
Þ Ð1 + Ð2 = Ð2 + Ð3 + Ð4
Þ Ð1 = Ð3 + Ð4
Q.E.D.
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5. Theorem 4: If two sides of a triangle are equal in measure, then the angles
opposite these sides are equal in measure. a b c d 3 4
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 = Ð ………… 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.
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6. Theorem 5: The opposite sides and opposite angles of a parallelogram are respectively equal in measure.
Given: Parallelogram abcd c b a d 3
To Prove: |ab| = |cd| and |ad| = |bc|
and Ðabc = Ðadc 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
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7. Theorem 6: A diagonal bisects the area of a parallelogramx
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
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Q.E.D

8. 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. a b c o
To Prove: | Ðboc | = 2 | Ðbac | r 2 5
Construction: Join a to o and extend to r
Proof: In the triangle aob 1 4 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 |
Þ | Ðboc | = 2 | Ðbac |
Q.E.D
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9. Theorem 8: A line through the centre of a circle perpendicular to a chord
bisects the chord. L
90 o o a b r
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
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10. Theorem 9: If two triangles are equiangular, the lengths of the corresponding sides are in proportion.
Given: Two Triangles with equal angles
To Prove:
|df|
|ac| =
|de|
|ab|
|ef|
|bc|
Construction: On ab mark off ax equal in length to de.
On ac mark off ay equal in length to df a c b d f e 1 2 3
Proof: Ð1 = Ð4
Þ [xy] is parallel to [bc]
|ay|
|ac| =
|ax|
|ab| Þ
As xy is parallel to bc x y 4 5
|df|
|ac| =
|de|
|ab|
Similarly
|ef|
|bc|
Q.E.D.
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11. 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.
Given: Triangle abc
To Prove: a2 + b2 = c2 a b c a b c
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. a b c 3 a b c 4 1 2
Must prove that it is a square. i.e. Show that │∠1 │= 90o
│∠1│+ │∠2│ =│∠3│+│∠4│ (external angle…)
But │∠2│=│∠3│ (Congruent triangles)
⇒│∠1│=│∠4│= 90o QED
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