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Theorems Leaving Certificate - Ordinary level Developed by Pádraic Kavanagh.

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Presentation on theme: "Theorems Leaving Certificate - Ordinary level Developed by Pádraic Kavanagh."— Presentation transcript:

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2 Theorems Leaving Certificate - Ordinary level Developed by Pádraic Kavanagh

3 Click on the theorem you what to revise! Theorem 1 Theorem 7 Theorem 6 Theorem 3 Theorem 4 Theorem 5 Theorem 8 Theorem 10 Theorem 9 Theorem 2 Quit

4 Proof: = Straight line 1 = 4 and 2 = 5 Alternate angles = = Q.E.D. 45 Given:Triangle 12 3 Construction:Draw line through 3 parallel to the base Theorem 1:The sum of the degree measures of the angles of a triangle is To Prove: = QuitMenu

5 Given:Parallelogram abcd cb ad Construction:Diagonal |ac| Proof: 1 = 4 Alternate angles |ac| = Given 2 = 3Alternate angles ASA abc = acd |ab| = |cd| and |ad| = |bc| Q.E.D. Theorem 2: The opposite sides of a parallelogram have equal lengths. To Prove:|ab| = |cd| and |ad| = |bc| QuitMenu

6 Given:Diagram as shown with |ab| = |bc| Proof: 1 = 2 Verticially opposite |ab| = |bc|Given 3 = 4Alternate angles ASA dab = bec |db| = |be| Q.E.D. a b c d e Construction:Another transversal through b. 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. To Prove:|db| = |be| QuitMenu

7 Given:Diagram as shown with line |xy| parallel to base. xy a b c Proof:|ax | is divided in m equal parts |ay| is also |xb| is divided in n equal parts |yc| is alsoTheorem 2 m equal parts n equal parts |yc| |ay| = |xb| |ax | Construction:Draw in division lines Q.E.D. 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. To Prove: |yc| |ay| = |xb| |ax | QuitMenu

8 Given:Two Triangles with equal angles Proof: def mapped onto axy 1 = 4 [xy] is parallel to [bc] Construction: Map def onto axy, draw in [xy] |df | |ac| = |de| |ab| Similarly |ef | |bc| = Q.E.D. a cb d fe xy 45 |ay| |ac| = |ax| |ab| Theorem 3 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. To Prove: |df | |ac| = |de| |ab| |ef | |bc| = QuitMenu

9 Given:Triangle abc Proof: Area of large sq. = area of small sq. + 4(area ) (a + b) 2 = c 2 + 4(½ab) a2 a2 + 2ab +b 2 = c 2 + 2ab a2 a2 + b2 b2 = c2c2 Q.E.D. a b c a b c a b c a b c Construction:Three right angled triangles as shown 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) To Prove:a 2 + b 2 = c 2 QuitMenu

10 Given:Triangle abc with |ac| 2 = |ab| 2 + |bc| 2 To Prove:To prove 1 = 90 0 Proof:|ac| 2 = |ab| 2 + |bc| 2 given |ac| 2 = |de| 2 + |ef| 2 from construction |ac| 2 = |df| 2 = |de| 2 + |ef| 2 |ac| = |df| SSS abc = def 1 = 90 0 Q.E.D. e d f a bc 1 2 Construction:Triangle def with|de| = |ab| |ef | = |bc| & 2 = 90 0 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) QuitMenu

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

12 Given:Triangle abc Construction:Draw [bd] such that |ab| = |ad| a b c d Proof: 1 = 2As |ab|= |ad| > 2 But 2 > > 4 abc > acb and the greater angle is opposite the longer side. Q.E.D 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. To Prove: abc > acb QuitMenu

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

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