Question

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

Answer 1

So, here I go:
First of I'll list up the hypothesis, which is the given information:
(H) ABC isoceles.
(T) AX = BX


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:
\[Ax=Bx\]

Answer 2

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

Answer 3

show they are equal too for isoc

Answer 4

: )

Answer 5

ill allow the usage of AAS SSS and SAS only

Answer 6

the ancient greeks proved this 2000 years ago :)

Answer 7

yup

Answer 8

using tan -.- cheater face

Answer 9

thats a high end function

Answer 10

and so show the base angles of isosceles are equal we can do