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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
    • one year ago
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    |dw:1435253588480:dw|

  2. Owlcoffee
    • one year ago
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    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
    • one year ago
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    the two base angles, say α, are equal so tanα = h/b1 = h/b2 where b1 + b2 = base therefore b1 = b2

  4. dan815
    • one year ago
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    show they are equal too for isoc

  5. dan815
    • one year ago
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    : )

  6. dan815
    • one year ago
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    ill allow the usage of AAS SSS and SAS only

  7. alekos
    • one year ago
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    the ancient greeks proved this 2000 years ago :)

  8. dan815
    • one year ago
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    yup

  9. dan815
    • one year ago
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    using tan -.- cheater face

  10. dan815
    • one year ago
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    thats a high end function

  11. dan815
    • one year ago
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    |dw:1435255353420:dw|

  12. dan815
    • one year ago
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    and so show the base angles of isosceles|dw:1435255424617:dw| are equal we can do

  13. dan815
    • one year ago
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    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
    • one year ago
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    I really like your method Dan, Ill have that in my repertoire.

  15. dan815
    • one year ago
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    if u like these kind of examples i have a fun one for u

  16. dan815
    • one year ago
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    you can try to solve this problem with only geometry its a famous one dont look it up xD

  17. Owlcoffee
    • one year ago
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    Give me your best shot.

  18. dan815
    • one year ago
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    |dw:1435255751221:dw|

  19. dan815
    • one year ago
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    |dw:1435255859970:dw|

  20. dan815
    • one year ago
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    |dw:1435255900735:dw|

  21. dan815
    • one year ago
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    here u go :)

  22. dan815
    • one year ago
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    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
    • one year ago
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    you must draw water from the river bank and go to the next city in the shortest route possible

  24. dan815
    • one year ago
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    is the question clear?

  25. Owlcoffee
    • one year ago
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    Yes

  26. dan815
    • one year ago
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    okay let a = 5, b= 10, L = 20 find point C|dw:1435256121850:dw|

  27. Owlcoffee
    • one year ago
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    |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
    • one year ago
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    |dw:1435260599276:dw|

  29. dan815
    • one year ago
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    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
    • one year ago
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    |dw:1435260688944:dw|

  31. dan815
    • one year ago
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    isnt it pretty :)

  32. Owlcoffee
    • one year ago
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    Wow, much simpler than I thought. And you used euclidean geometry... Reflections...

  33. dan815
    • one year ago
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    it's one of the most elegant solutions from history

  34. Owlcoffee
    • one year ago
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    With no wonder. Very straight and elegantly solved.

  35. dan815
    • one year ago
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    haha its looks very simple once u see it, but boy ill tell you, its very rare someone would find this solution

  36. dan815
    • one year ago
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    especially nowadays when everyone just resorts to calculus

  37. dan815
    • one year ago
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    this one can be solved as a minimization problem, and gets all messy

  38. dan815
    • one year ago
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    and an interesting thing comes which is that these angles are equal

  39. dan815
    • one year ago
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    |dw:1435261128090:dw|

  40. dan815
    • one year ago
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    so the point always such that those angles are equal, so these 2 triangles are similiar

  41. dan815
    • one year ago
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    and the main triangle here can be draw too to solve completely

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

  43. dan815
    • one year ago
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    |dw:1435261208966:dw|

  44. dan815
    • one year ago
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    |dw:1435261234039:dw|

  45. Owlcoffee
    • one year ago
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    proportionality. I really need to dig back to geometry, I am starting to get rusty.

  46. dan815
    • one year ago
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    theres multiple ways to solve it but this is the mains solution |dw:1435261434618:dw|

  47. dan815
    • one year ago
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    once u draw this pic u are pretty much done technically, as u just managed to draw shortest line

  48. dan815
    • one year ago
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    in books, they usually stop there

  49. Owlcoffee
    • one year ago
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    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
    • one year ago
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    there an extension to this, which is in higher dimension using same idea

  51. dan815
    • one year ago
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    or like if u have to hit 2 lines

  52. Owlcoffee
    • one year ago
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    In 3 dimensions?

  53. dan815
    • one year ago
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    |dw:1435261603447:dw|

  54. dan815
    • one year ago
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    like this one, very similar , but now u have to hit 2 lines before going from town a to b

  55. dan815
    • one year ago
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    maybe u can find this one :) and get back to me tomorrow

  56. Owlcoffee
    • one year ago
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    I will try, I believe you have many questions to answer nowdays :x

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