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

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Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1So, 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: \[Ax=Bx\]

alekos
 one year ago
Best ResponseYou've already chosen the best response.0the two base angles, say α, are equal so tanα = h/b1 = h/b2 where b1 + b2 = base therefore b1 = b2

dan815
 one year ago
Best ResponseYou've already chosen the best response.1show they are equal too for isoc

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ill allow the usage of AAS SSS and SAS only

alekos
 one year ago
Best ResponseYou've already chosen the best response.0the ancient greeks proved this 2000 years ago :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1using tan . cheater face

dan815
 one year ago
Best ResponseYou've already chosen the best response.1thats a high end function

dan815
 one year ago
Best ResponseYou've already chosen the best response.1and so show the base angles of isoscelesdw:1435255424617:dw are equal we can do

dan815
 one year ago
Best ResponseYou've already chosen the best response.1still kind of cheating because im using SSS , but it would be really nice to see without that, or show that proof too

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1I really like your method Dan, Ill have that in my repertoire.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1if u like these kind of examples i have a fun one for u

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you can try to solve this problem with only geometry its a famous one dont look it up xD

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1Give me your best shot.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1think 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
 one year ago
Best ResponseYou've already chosen the best response.1you must draw water from the river bank and go to the next city in the shortest route possible

dan815
 one year ago
Best ResponseYou've already chosen the best response.1okay let a = 5, b= 10, L = 20 find point Cdw:1435256121850:dw

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1dw: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: \[m=\frac{ 0x_c }{ 50 }=\frac{ x_c }{ 5 }\] \[t)(y5)=(\frac{ x_c }{ 5 })(x)\] \[t)xx_c+2y25=0\] And If I find the other line which is composed by C and B: \[m=\frac{ 20x_c }{ 100 }\] \[r)(y10)=(\frac{ 20x_c }{ 10 })(x20)\] \[r)10y100=20x400xx_c20x_c\] \[r)(x_c20)x+10y+(20x_c+300)=0\] So, the intersection of t) , r) and the line y=0 must give me as a result the xcoordinate of the point C in my reference system.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1what 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

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1Wow, much simpler than I thought. And you used euclidean geometry... Reflections...

dan815
 one year ago
Best ResponseYou've already chosen the best response.1it's one of the most elegant solutions from history

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1With no wonder. Very straight and elegantly solved.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1haha its looks very simple once u see it, but boy ill tell you, its very rare someone would find this solution

dan815
 one year ago
Best ResponseYou've already chosen the best response.1especially nowadays when everyone just resorts to calculus

dan815
 one year ago
Best ResponseYou've already chosen the best response.1this one can be solved as a minimization problem, and gets all messy

dan815
 one year ago
Best ResponseYou've already chosen the best response.1and an interesting thing comes which is that these angles are equal

dan815
 one year ago
Best ResponseYou've already chosen the best response.1so the point always such that those angles are equal, so these 2 triangles are similiar

dan815
 one year ago
Best ResponseYou've already chosen the best response.1and the main triangle here can be draw too to solve completely

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1Well yes, they are equal because of vertical angles, and since one is the reflection, the third must be equal as well.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1proportionality. I really need to dig back to geometry, I am starting to get rusty.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1theres multiple ways to solve it but this is the mains solution dw:1435261434618:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.1once u draw this pic u are pretty much done technically, as u just managed to draw shortest line

dan815
 one year ago
Best ResponseYou've already chosen the best response.1in books, they usually stop there

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1there an extension to this, which is in higher dimension using same idea

dan815
 one year ago
Best ResponseYou've already chosen the best response.1or like if u have to hit 2 lines

dan815
 one year ago
Best ResponseYou've already chosen the best response.1like this one, very similar , but now u have to hit 2 lines before going from town a to b

dan815
 one year ago
Best ResponseYou've already chosen the best response.1maybe u can find this one :) and get back to me tomorrow

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1I will try, I believe you have many questions to answer nowdays :x
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