2. Menu Theorem 1 Vertically opposite angles are equal in measure. Theorem 2 The measure of the three angles of a triangle sum to 180 o. 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.

3. 1 2 3 4 0 180 90 45 135 0 180 90 45 135 Theorem 1: Vertically opposite angles are equal in measure To Prove: To Prove:  1 =  3 and  2 =  4 Proof:  1 +  2 = 180 0 ………….. Straight line  2 +  3 = 180 0 ………….. Straight line   1 +  2 =  2 +  3  1 =  3 Similarly  2 =  4 Q.E.D. Menu

4. Proof:  3  3 +  4  4 +  5  5 = 180 0 Straight line  1  1 =  4  4 and  2  2 = 5555Alternate angles   3  3 +  1  1 +  2  2 = 180 0  1  1 +  2  2 +  3  3 = 180 0 Q.E.D. 45 Given: Given:Triangle 12 3 Construction: Construction:Draw line through  3 parallel to the base Theorem 2:The measure of the three angles of a triangle sum to 180 0. To Prove: To Prove:  1 +  2 +  3 = 180 0 Menu

5. 0 180 90 45 135 Theorem 3: An exterior angle of a triangle equals the sum of the two interior opposite angles in measure. opposite angles in measure. To Prove: To Prove:  1 =  3 +  4 Proof:  1 +  2 = 180 0 ………….. Straight line  2 +  3 +  4 = 180 0 ………….. Triangle.  1 +  2 =  2 +  3 +  4  1 =  3 +  4 Q.E.D. 1 2 3 4 Menu

6. Theorem 4: If two sides of a triangle are equal in measure, then the angles opposite these sides are equal in measure. opposite these sides are equal in measure. To Prove: To Prove:  1 =  2 Proof: In the triangle abd and the triangle adc |ab| = |ac| ………….. Given.  ……….. SAS = SAS.  The triangle abd is congruent to the triangle adc  ……….. SAS = SAS. Q.E.D. 1 2 3 4 a b c Given: |ab| = |ac| Given: Triangle abc with |ab| = |ac| Construction:  bac Construction:Construct ad the bisector of  bac d  3 =  4 ………….. Construction |ad| = |ad| ………….. Common Side.   1 =  2 Menu

7. 2 3 1 4 Given: Given:Parallelogram abcd cb ad Construction: Construction: Draw the diagonal |ac| Theorem 5: The opposite sides and opposite angles of a parallelogram are respectively equal in measure. To Prove: To Prove:|ab| = |cd| and |ad| = |bc|  abc =  adc and  abc =  adc Proof:In the triangle abc and the triangle adc  1 =  4 …….. Alternate angles |ac| = |ac| …… Common  2 =  3 ……… Alternate angles  ……… ASA = ASA.  The triangle abc is congruent to the triangle adc  ……… ASA = ASA.  |ab| = |cd| and |ad| = |bc|  abc =  adc and  abc =  adc Q.E.D Menu

8. Given: Given:Parallelogram abcd Construction: Construction: Draw perpendicular from b to ad Theorem 6: A diagonal bisects the area of a parallelogram To Prove: To Prove:Area of the triangle abc = Area of the triangle adc Proof:Area of triangle adc = ½ |ad| x |bx| Area of triangle abc = ½ |bc| x |bx| As |ad| = |bc| …… Theorem 5   The diagonal ac bisects the area of the parallelogram Q.E.D b c ad x Area of triangle adc = Area of triangle abc Menu

9. 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. measure of the angle at the circumference standing on the same arc. To Prove:|  boc | = 2 |  bac | To Prove: |  boc | = 2 |  bac | Construction: Construction: Join a to o and extend to r r Proof:In the triangle aob a b c o 1 3 2 4 5 | oa| = | ob | …… Radii  |  2 | = |  3 | …… Theorem 4 |  1 | = |  2 | + |  3 | …… Theorem 3 |  1 | = |  2 | + |  3 | …… Theorem 3  |  1 | = |  2 | + |  2 |  |  1 | = 2|  2 |  Similarly |  4 | = 2|  5 |  |  boc | = 2 |  bac | Q.E.D Menu

10. Theorem 8: A line through the centre of a circle perpendicular to a chord bisects the chord. bisects the chord. Given: Given:A circle with o as centre and a line L perpendicular to ab. To Prove:| ar | = | rb | To Prove: | ar | = | rb | L 90 o o a b r Proof:In the triangles aor and the triangle orb Construction: Construction: Join a to o and o to b |ao| = |ob| ………….. Radii.  aro =  orb …………. 90 o |or| = |or| ………….. Common Side.  ……… RSH = RSH.  The triangle aor is congruent to the triangle orb  ……… RSH = RSH.  |ar| = |rb| Q.E.D Menu

11. Given: Given: Two Triangles with equal angles Proof:  1  1 = 4444 [xy] is parallel to [bc] Construction: Construction: On ab mark off ax equal in length to de. On ac mark off ay equal in length to df |df| |ac| = |de| |ab| Similarly |ef| |bc| = Q.E.D. a cb d fe 1 2 3 13 2 xy 45 |ay| |ac| = |ax| |ab|  As xy is parallel to bc Theorem 9:If two triangles are equiangular, the lengths of the corresponding sides are in proportion. To Prove: |df| |ac| = |de| |ab| |ef| |bc| = Menu

12. Given: Given:Triangle abc Proof:Area of large sq. = area of small sq. + 4(area )))) (a + b) 2 b) 2 = c 2 c 2 + 4(½ab) a2 a2 a2 a2 + 2ab +b 2 +b 2 = c 2 c 2 + 2ab a2 a2 a2 a2 + b2 b2 b2 b2 = c2c2c2c2Q.E.D. a b c a b c a b c a b c Construction: Construction:Three right angled triangles as shown 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. To Prove: To Prove:a 2 + b 2 = c 2 Menu Must prove that it is a square. i.e. Show that │ ∠1 │ = 90 o 1 4 3 2 │ ∠1 │ + │ ∠2 │ =│ ∠3 │+│ ∠4 │ (external angle…) But │ ∠2 │=│ ∠3 │ (Congruent triangles) ⇒ │ ∠1 │=│ ∠4 │= 90 o QED