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Triangle Proof by Kathy McDonald section 3.1 #7. Prove: When dividing each side of an equilateral triangle.

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Presentation on theme: "Triangle Proof by Kathy McDonald section 3.1 #7. Prove: When dividing each side of an equilateral triangle."— Presentation transcript:

1 Triangle Proof by Kathy McDonald section 3.1 #7

2 Prove: When dividing each side of an equilateral triangle

3 into n segments

4 then connecting the division points with all possible segments parallel to the original sides, n² small triangles are created.

5 Proof by induction: Let S = {n  N: f(n) = n²}

6 1 Show 1  S: f(n) =n² f(1) = 1 = 1²

7 Show 2  S: when dividing each side into 2 segments

8 and connecting division points as described,

9

10

11 4 small triangles are created.

12 f(n) =n² f(2) = 4 = 2²

13 Show 3  S: when dividing each side into 3 segments

14 and connecting division points as described,

15

16

17 9 small triangles are created.

18 f(n) =n² f(3) = 9 = 3²

19 Assume n  S. Assume when dividing each side into n segments and connecting division points as described, n² small triangles are created. Assume f(n) = n².

20 Show n+1  S. Show when dividing each side into n+1 segments and connecting division points as described, (n+1)² small triangles are created. Show f(n+1) = (n+1)².

21 Consider a divided triangle with n segments on each side.

22 When a segment equal in size to the n segments is added to each side

23 and those endpoints are connected,

24 a space is created at the bottom of the original triangle. Also, a new, bigger equilateral triangle has been created.

25 This new, bigger triangle has n+1 segments on each side. n segments + 1 segment

26 Now, the parallel dividing lines are extended down to the base of the new, bigger triangle.

27 More small triangles are created.

28 The n segments of the base of the original triangle

29 correspond to n bases of the new, small triangles created.

30 Also, the n+1 segments of the base of the new, bigger triangle

31 correspond to n+1 bases of the new, small triangles.

32 So, n+(n+1) bases

33 correspond to n+(n+1) new, small triangles

34 By assumption, the original triangle has n segments on each side And n² small triangles inside.

35 By adding 1 segment to each side of this triangle, n + (n+1) small triangles are added.

36 The total small triangles of the new, bigger triangle is:

37 =n²+2n+1 =(n+1)(n+1) n² + n +(n+1) = (n+1)²

38 This shows n+1  S. By induction, S  N.

39 Dwight says, “that’s it.”


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