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dan815 one year ago Prove height of isosceles intersects the midpoint of the base @owlcoffee

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1. dan815

|dw:1435253588480:dw|

2. Owlcoffee

So, here I go: First of I'll list up the hypothesis, which is the given information: (H) ABC isoceles. (T) AX = BX |dw:1435253753194:dw| To begin, I'll take the very definition of isoceles triangle: $(1) CA=CB$ and : $(2) <xAC=<xBC$ And by reflexive property: $(3) Cx=Cx$ though people rush to use SAS here, it's not quite correct, because I don't know rigurously if <xCA and <xCB are equal, and we can simply use the theorem of the sum of angles in a triangle. But first, use the definition of height: $Cx \perp AB$ therefore: $<AxC=90$ $<BxC=90$ and by transitive: $(4)<AxC = < BxC$ $<xAC + <AxC + < xCA=180$ and for the other: $<xBC+<CxB+<BCx=180$ By the transitive property I can conclude: $<xAC + <AxC + < xCA=<xBC+<CxB+<BCx$ And because of (2) and (4): $(5)<xCA=<BCx$ And now, using (1), (3) and (5) as premises, and the axiom of SAS, I can conclude: $(6)\triangle AxC= \triangle BxC$ Then, by (6) concluding: $Ax=Bx$

3. alekos

the two base angles, say α, are equal so tanα = h/b1 = h/b2 where b1 + b2 = base therefore b1 = b2

4. dan815

show they are equal too for isoc

5. dan815

: )

6. dan815

ill allow the usage of AAS SSS and SAS only

7. alekos

the ancient greeks proved this 2000 years ago :)

8. dan815

yup

9. dan815

using tan -.- cheater face

10. dan815

thats a high end function

11. dan815

|dw:1435255353420:dw|

12. dan815

and so show the base angles of isosceles|dw:1435255424617:dw| are equal we can do

13. dan815

still kind of cheating because im using SSS , but it would be really nice to see without that, or show that proof too

14. Owlcoffee

I really like your method Dan, Ill have that in my repertoire.

15. dan815

if u like these kind of examples i have a fun one for u

16. dan815

you can try to solve this problem with only geometry its a famous one dont look it up xD

17. Owlcoffee

Give me your best shot.

18. dan815

|dw:1435255751221:dw|

19. dan815

|dw:1435255859970:dw|

20. dan815

|dw:1435255900735:dw|

21. dan815

here u go :)

22. dan815

think of the line as water, the distance from the city A to the water is a and City B to water is b and the horizontal distance from city A to City B is L

23. dan815

you must draw water from the river bank and go to the next city in the shortest route possible

24. dan815

is the question clear?

25. Owlcoffee

Yes

26. dan815

okay let a = 5, b= 10, L = 20 find point C|dw:1435256121850:dw|

27. Owlcoffee

|dw:1435256155530:dw| Let a ortonormated reference system exist, such way that the segment "a" intersection with the line "L" defines the origin. Therefore: $A(0,5)$ $B(20, 10)$ $C(x_c , y_c)$ But since C belongs in my actual x_axis, then: $C(x_c, 0)$ So if I find the line from C to A, then: $m=\frac{ 0-x_c }{ 5-0 }=\frac{ -x_c }{ 5 }$ $t)(y-5)=(\frac{ -x_c }{ 5 })(x)$ $t)xx_c+2y-25=0$ And If I find the other line which is composed by C and B: $m=\frac{ 20-x_c }{ 10-0 }$ $r)(y-10)=(\frac{ 20-x_c }{ 10 })(x-20)$ $r)10y-100=20x-400-xx_c-20x_c$ $r)(x_c-20)x+10y+(20x_c+300)=0$ So, the intersection of t) , r) and the line y=0 must give me as a result the x-coordinate of the point C in my reference system.

28. dan815

|dw:1435260599276:dw|

29. dan815

what you do is reflect a across the horizontal line, and connect point b to reflection of a, now this is a straight line and you will see that any other point along this river will yield a line bigger

30. dan815

|dw:1435260688944:dw|

31. dan815

isnt it pretty :)

32. Owlcoffee

Wow, much simpler than I thought. And you used euclidean geometry... Reflections...

33. dan815

it's one of the most elegant solutions from history

34. Owlcoffee

With no wonder. Very straight and elegantly solved.

35. dan815

haha its looks very simple once u see it, but boy ill tell you, its very rare someone would find this solution

36. dan815

especially nowadays when everyone just resorts to calculus

37. dan815

this one can be solved as a minimization problem, and gets all messy

38. dan815

and an interesting thing comes which is that these angles are equal

39. dan815

|dw:1435261128090:dw|

40. dan815

so the point always such that those angles are equal, so these 2 triangles are similiar

41. dan815

and the main triangle here can be draw too to solve completely

42. Owlcoffee

Well yes, they are equal because of vertical angles, and since one is the reflection, the third must be equal as well.

43. dan815

|dw:1435261208966:dw|

44. dan815

|dw:1435261234039:dw|

45. Owlcoffee

proportionality. I really need to dig back to geometry, I am starting to get rusty.

46. dan815

theres multiple ways to solve it but this is the mains solution |dw:1435261434618:dw|

47. dan815

once u draw this pic u are pretty much done technically, as u just managed to draw shortest line

48. dan815

in books, they usually stop there

49. Owlcoffee

Yes, indeed. I think when I used analytical geometry i limited myself to finding the point C. And I can understand why people resort to calculus, you can translate it to a function and find the minima.

50. dan815

there an extension to this, which is in higher dimension using same idea

51. dan815

or like if u have to hit 2 lines

52. Owlcoffee

In 3 dimensions?

53. dan815

|dw:1435261603447:dw|

54. dan815

like this one, very similar , but now u have to hit 2 lines before going from town a to b

55. dan815

maybe u can find this one :) and get back to me tomorrow

56. Owlcoffee

I will try, I believe you have many questions to answer nowdays :x

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