Question

proof and reasoning

Answer 1

http://prntscr.com/hk5a6c

Answer 2

@563blackghost

Answer 3

anyone..

Answer 4

@Vocaloid

Answer 5

heyo, let me take a look at it...

Answer 6

thnxx

Answer 7

this might take a while...i have done triangles but this seems a bit different from the ones i've done...

Answer 8

take your time

Answer 9

i think it needs like one of those ssa aas yknow rules

Answer 10

SSA is not a proper rule you can use btw

Answer 11

in any case
AB = AC by definition of isosceles triangle

Answer 12

XB = XC by definition of midpoint

Answer 13

then we want to say that <ABC = <ACB (I forget the reason)

in order to use SAS to say the triangles BAX and CAX are similar triangles

then CPCTC to say that <BAX = <CAX

Answer 14

hmmm

We are given that \(\bf{\triangle ABC}\) is Isoceles with Vertex \(\bf{\angle BAC}\) and that \(\bf{X}\) is the midpoint of \(\bf{BC}\).

Since we are told that the triangle is isoceles because of the vertex it would mean that \(\bf{AB=AC}\) due to def. of isosceles triangle.

Since X is the midpoint of \(\bf{BC}\) this means that the segment is split in the middle, this saying that \(\bf{BX=BC}\)..


....and @Angle already said this....

Answer 15

well then...

I would agree with the CPCTC...

Answer 16

@563blackghost I like the way you explained it better :)

I missed reason for one of the things, do you know what it is? I can't think of it

Answer 17

Reflexive probably? for AX=AX?

Answer 18

no...wait that would be after...

Answer 19

<ABC = <ACB

it's the base angles of the isosceles triangle are equal
I'm not sure what the specific rule is called or if you're allowed to use a rule like that in the proof

Answer 20

Thank you both sooooooooooooo much!!

Answer 21

I wouldn't know how to apply that as part of the proof...i feel like im missing something...

Answer 22

Given X is the midpoint of BC
Given

Answer 23

@Logic007
As I mentioned in my first comment, SSA cannot be used to prove triangle congruence

Answer 24

Which means that BXA is right and CXA is right too. Meaning AX=AX thus having that angle BAX is congruent to CAX by CPTCTC...no??

Answer 25

@563blackghost
CPCTC doesn't work for SSA or SSS

Answer 26

We proved that BX=CA, and we can see that BXA is right and CXA is right too meaning that BXA=CXA. They share a segment being AX=AX by reflexive. We proved triangle BXA= triangle CXA by SAS...corresponding sides are equal to each other so angle BAX = angle CAX...

hmmmmmm

Answer 27

no...i cant just assume right angle...

darn

Answer 28

ohhh
yes
that is another way to prove it

but saying the perpendicular bisector of BC goes through X and A is a little difficult to prove; although it should have a name to this rule

Answer 29

well for X to intersect vertex A would be definition of a bisector....

Answer 30

Im thinking of saying BA=CA and BX=CX thus \({\angle XBA=\angle XCA}\) but i dont think that works....

Answer 31

Are you sure this is the whole picture, cause the pic looks a little cut off (it could be showing right angles....) just asking @Sarah10

Answer 32

yes

Answer 33

ok

Answer 34

i swear i have been trying to keep up .. can someone please summarize it if you have time

Answer 35

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Angle
@563blackghost
CPCTC doesn't work for SSA or SSS
\(\color{#0cbb34}{\text{End of Quote}}\)

This does work. Did some researching.

Answer 36

What reason did you find for \(\bf{\angle ABX = \angle ACX}\)? Just wondering.

Answer 37

https://mathbitsnotebook.com/Geometry/CongruentTriangles/CTtriangleMethods.html

Answer 38

XD i didn't see that...ah ok

Answer 39

http://www.mathwarehouse.com/geometry/congruent_triangles/isosceles-triangle-theorems-proofs.php
Base Angles Theorem for Isosceles Triangles

Answer 40

writing this one down....