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

- dan815

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

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- katieb

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- dan815

|dw:1435253588480:dw|

- 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)

- alekos

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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## More answers

- dan815

show they are equal too for isoc

- dan815

: )

- dan815

ill allow the usage of AAS SSS and SAS only

- alekos

the ancient greeks proved this 2000 years ago :)

- dan815

yup

- dan815

using tan -.- cheater face

- dan815

thats a high end function

- dan815

|dw:1435255353420:dw|

- dan815

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

- dan815

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

- Owlcoffee

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

- dan815

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

- dan815

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

- Owlcoffee

Give me your best shot.

- dan815

|dw:1435255751221:dw|

- dan815

|dw:1435255859970:dw|

- dan815

|dw:1435255900735:dw|

- dan815

here u go :)

- 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

- dan815

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

- dan815

is the question clear?

- Owlcoffee

Yes

- dan815

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

- 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.

- dan815

|dw:1435260599276:dw|

- 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

- dan815

|dw:1435260688944:dw|

- dan815

isnt it pretty :)

- Owlcoffee

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

- dan815

it's one of the most elegant solutions from history

- Owlcoffee

With no wonder.
Very straight and elegantly solved.

- dan815

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

- dan815

especially nowadays when everyone just resorts to calculus

- dan815

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

- dan815

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

- dan815

|dw:1435261128090:dw|

- dan815

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

- dan815

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

- Owlcoffee

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

- dan815

|dw:1435261208966:dw|

- dan815

|dw:1435261234039:dw|

- Owlcoffee

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

- dan815

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

- dan815

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

- dan815

in books, they usually stop there

- 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.

- dan815

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

- dan815

or like if u have to hit 2 lines

- Owlcoffee

In 3 dimensions?

- dan815

|dw:1435261603447:dw|

- dan815

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

- dan815

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

- Owlcoffee

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

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