Download presentation

1
**Leaving Certificate - Ordinary level**

Theorems Leaving Certificate - Ordinary level Developed by Pádraic Kavanagh

2
**Click on the theorem you what to revise!**

Quit

3
**Theorem 1: The sum of the degree measures of the angles of a triangle is 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. Quit Menu

4
**Theorem 2: The opposite sides of a parallelogram have equal lengths.**

Given: Parallelogram abcd To Prove: |ab| = |cd| and |ad| = |bc| Construction: Diagonal |ac| Proof: Ð1 = Ð4 Alternate angles |ac| = |ac| Given Ð2 = Ð3 Alternate angles ASA Þ D abc = D acd Þ |ab| = |cd| and |ad| = |bc| Q.E.D. c b a d 2 3 1 4 Quit Menu

5
Theorem 3: If three parallel lines make intercepts of equal length on a transversal, then they will also make equal length on any other transversal. Given: Diagram as shown with |ab| = |bc| To Prove: |db| = |be| Construction: Another transversal through b. Proof: Ð1 = Ð2 Verticially opposite |ab| = |bc| Given Ð3 = Ð4 Alternate angles ASA Þ D dab = D bec Þ |db| = |be| Q.E.D. a b c d e 1 2 3 4 Quit Menu

6
Theorem 4: A line which is parallel to one side-line of a triangle, and cuts a second side, will cut the third side in the same proportion as the second. Given: Diagram as shown with line |xy| parallel to base. To Prove: |yc| |ay| = |xb| |ax | x y a b c Construction: Draw in division lines m equal parts n equal parts Proof: |ax | is divided in m equal parts Þ |ay| is also |xb| is divided in n equal parts Þ |yc| is also Theorem 2 |yc| |ay| = |xb| |ax | Q.E.D. Quit Menu

7
Theorem 5: If the three angles of one triangle have degree-measures equal, respectively, to the degree-measures of the angles of a second triangle, then the lengths of the corresponding sides of the two triangles are proportional. Given: Two Triangles with equal angles To Prove: |df | |ac| = |de| |ab| |ef | |bc| a c b d f e 1 2 3 Construction: Map D def onto D axy, draw in [xy] Proof: D def mapped onto D axy Ð1 = Ð4 Þ [xy] is parallel to [bc] x y 4 5 |ay| |ac| = |ax| |ab| Þ Theorem 3 |df | |ac| = |de| |ab| Similarly |ef | |bc| Q.E.D. Quit Menu

8
**a b c a b c a b c a b c Quit Menu**

Theorem 6: 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 lengths of the other two sides. (Pythagoras) 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 a b c Quit Menu

9
Theorem 7: If the square of the length of one side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle has a right-angle and this is opposite the longest side. (Converse of Pythagoras’ Theorem) Given: Triangle abc with |ac|2 = |ab|2 + |bc|2 Construction: Triangle def with |de| = |ab| |ef | = |bc| & Ð2 = 900 a b c 1 e d f To Prove: To prove Ð1 = 900 Proof: |ac|2 = |ab|2 + |bc|2 given |ac|2 = |de|2 + |ef|2 from construction |ac|2 = |df|2 |df|2 = |de|2 + |ef|2 Þ |ac| = |df| SSS Þ D abc = D def Ð1 = 900 Q.E.D. 2 Quit Menu

10
**Theorem 8:. The products of the lengths of the sides of a triangle by**

Theorem 8: The products of the lengths of the sides of a triangle by the corresponding altitudes are equal. Given: Triangle abc To Prove: |bc||ad| = |ac||be| Construction: Altitudes [be] and [ad] Proof: Compare D bce to D acd 4 5 1 c d a 3 2 b e a b c 4 5 1 3 2 d e Ð1 = Ð1 Ð2 = Ð4 Þ Ð3 = Ð5 Similar triangles |ad| |be| = |ac| |bc| Þ |bc||ad| = |ac||be| Q.E.D. Quit Menu

11
Theorem 9: If lengths of two sides of a triangle are unequal, then the degree-measures of the angles opposite to them are unequal, with the greater angle opposite to the longer side. Given: Triangle abc To Prove: Ðabc > Ðacb Construction: Draw [bd] such that |ab| = |ad| Proof: Ð1 = Ð2 As |ab|= |ad| Þ Ð1 + Ð3 > Ð2 But Ð2 > Ð4 Þ Ð1 + Ð3 > Ð4 Þ Ðabc > Ðacb and the greater angle is opposite the longer side. Q.E.D a b c d 3 1 2 4 Quit Menu

12
Theorem 10: The sum of the lengths of any two sides of a triangle is greater than that of the third side. Given: Triangle abc To Prove: |ba| + |ac| > |bc| Construction: Draw D acd such that |ad| = |ac| Proof: In D acd |ad| = |ac| Ð1 = Ð2 Þ Ð1 + Ð3 > Ð2 Þ |bd| > |bc| But |bd| = |ba| + |ad| |bd| = |ba| + |ac| as |ad| = |bc| Þ |ba| + |ac| > |bc| Q.E.D. 3 d 1 2 b a c Quit Menu

13
**YES NO Are you sure you want to quit**

The developer is not responsible for errors or omissions

Similar presentations

Presentation is loading. Please wait....

OK

Geometry Math 2. Proofs Lines and Angles Proofs.

Geometry Math 2. Proofs Lines and Angles Proofs.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

360 degree customer view ppt on mac Ppt on leadership in organizations Ppt on market friendly state political cartoon Ppt on statistics in maths lesson Ppt on different types of forests in the us Ppt on blood and its components Ppt on isobars and isotopes definition Ppt on ram and rom definition Microsoft office ppt online shopping Ppt on conservation of biodiversity