Prove height of isosceles intersects the midpoint of the base @owlcoffee

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Prove height of isosceles intersects the midpoint of the base @owlcoffee

Mathematics
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|dw:1435253588480:dw|
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)
the two base angles, say α, are equal so tanα = h/b1 = h/b2 where b1 + b2 = base therefore b1 = b2

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show they are equal too for isoc
: )
ill allow the usage of AAS SSS and SAS only
the ancient greeks proved this 2000 years ago :)
yup
using tan -.- cheater face
thats a high end function
|dw:1435255353420:dw|
and so show the base angles of isosceles|dw:1435255424617:dw| are equal we can do
still kind of cheating because im using SSS , but it would be really nice to see without that, or show that proof too
I really like your method Dan, Ill have that in my repertoire.
if u like these kind of examples i have a fun one for u
you can try to solve this problem with only geometry its a famous one dont look it up xD
Give me your best shot.
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here u go :)
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
you must draw water from the river bank and go to the next city in the shortest route possible
is the question clear?
Yes
okay let a = 5, b= 10, L = 20 find point C|dw:1435256121850:dw|
|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.
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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
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isnt it pretty :)
Wow, much simpler than I thought. And you used euclidean geometry... Reflections...
it's one of the most elegant solutions from history
With no wonder. Very straight and elegantly solved.
haha its looks very simple once u see it, but boy ill tell you, its very rare someone would find this solution
especially nowadays when everyone just resorts to calculus
this one can be solved as a minimization problem, and gets all messy
and an interesting thing comes which is that these angles are equal
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so the point always such that those angles are equal, so these 2 triangles are similiar
and the main triangle here can be draw too to solve completely
Well yes, they are equal because of vertical angles, and since one is the reflection, the third must be equal as well.
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proportionality. I really need to dig back to geometry, I am starting to get rusty.
theres multiple ways to solve it but this is the mains solution |dw:1435261434618:dw|
once u draw this pic u are pretty much done technically, as u just managed to draw shortest line
in books, they usually stop there
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.
there an extension to this, which is in higher dimension using same idea
or like if u have to hit 2 lines
In 3 dimensions?
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like this one, very similar , but now u have to hit 2 lines before going from town a to b
maybe u can find this one :) and get back to me tomorrow
I will try, I believe you have many questions to answer nowdays :x

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