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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.

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  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 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.

  2. 90 90 45 45 135 135 0 0 180 180 Theorem 1: Vertically opposite angles are equal in measure 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. Menu

  3. 4 5 3 1 2 Theorem 2: The measure of the three angles of a triangle sum to 1800 . 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 Menu

  4. 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. To Prove:Ð1 = Ð3 + Ð4 Proof:Ð1 + Ð2 = 1800 …………..Straight line Ð2 + Ð3 + Ð4 = 1800 ………….. Triangle. Þ Ð1 + Ð2 = Ð2 + Ð3 + Ð4 Þ Ð1 = Ð3 + Ð4 Q.E.D. Menu

  5. 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. 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. Menu

  6. b c a d Theorem 5: The opposite sides and opposite angles of a parallelogram are respectively equal in measure. 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 Menu

  7. c b a d x Theorem 6: A diagonal bisects the area of a parallelogram 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 Menu Q.E.D

  8. 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. 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 | Menu

  9. L a o r 90 o b Theorem 8: A line through the centre of a circle perpendicular to a chord bisects the chord. 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 Menu

  10. |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. 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. Menu

  11. 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. 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 = c2Q.E.D. 3 4 1 2 Must prove that it is a square. i.e. Show that │∠1 │= 90o │∠1│+ │∠2│ =│∠3│+│∠4│ (external angle…) ⇒│∠1│=│∠4│= 90o QED But │∠2│=│∠3│ (Congruent triangles) Menu

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