can some one help me?

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can some one help me?

Geometry
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https://media.glynlyon.com/g_geo_2012/9/q415a.gif
1 Attachment
Prove: In an equilateral triangle the three medians are equal.
what do you have so far?

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Other answers:

absolutely nothing ive been staring at the page for an hour
|dw:1439083430829:dw|
|dw:1439083442888:dw|
|dw:1439083487262:dw|
do you see how P is the midpoint of AB?
yes
they use the midpoint formula to find point P
that means you add up the x coordinates and divide by 2 |dw:1439083618310:dw|
make sense?
somewhat
I added the x coordinates 0 and 2a to get 2a then I divided 2a by 2 to get 'a'
|dw:1439083746502:dw|
ok got it
the same happens with the y coordinates |dw:1439083780104:dw|
ok
|dw:1439083807218:dw|
so hopefully you see how (a,0) was found
ya think so
the same idea is applied to find Q and R
feels like a proof... like verify (a,0) using A and B
okay
i think i understand.. thanks
|dw:1439083816461:dw|
for the other stuff, you use the distance formula \[\Large d = \sqrt{\left(x_{2}-x_{1}\right)^2+\left(y_{2}-y_{1}\right)^2}\]
kk
|dw:1439083949599:dw|
???
this assignment looks painful D: but I sort of see where they are going.
it is very painful i have 7 more
clicked close by accident.
so we need Q and R hmmm Q is (3a/2,b/2) R is (a/2, b/2)
lol i wish. It only lets me do that for quixs and test
no I said I accidentally closed the tab on my end XD
ohhhhh lol
so I couldn't reach you for a bit XD
if Q ( 3a/2, b/2) then there are points that can lead us to this. if R ( a/2,b/2) then there are points that lead to this result points A and C may lead us to R
A (0,0) C (a,b) |dw:1439084550670:dw|
okay
|dw:1439084709293:dw|
a HA! B AND C can make Q!
B(2a,0) C( a,b) (2a+a/2, 0+b/2) Q(3a/2,b/2)
|dw:1439084868914:dw|
|dw:1439084931274:dw|
ohh!!! i understand now thank you sooo much!!!!
I think midpoint formula and distance formula is bring used here.
QA hmm Q(3a/2, b/2) A (0,0)
|dw:1439085179111:dw|
\[\Large d = \sqrt{\left(x_{2}-x_{1}\right)^2+\left(y_{2}-y_{1}\right)^2} \] \[\Large d = \sqrt{(0-\frac{3a}{2})^2+(0-\frac{b}{2})^2} \] \[\Large d = \sqrt{(-\frac{3a}{2})^2+(-\frac{b}{2})^2} \] \[\Large d = \sqrt{(\frac{9a^2}{4})+\frac{b^2}{4}} \] if we let \[\large b = a \sqrt{3} \] \[\Large d = \sqrt{(\frac{9a^2}{4})+\frac{(a \sqrt{3})^2}{4}} \] \[\Large d = \sqrt{(\frac{9a^2}{4})+\frac{(3a^2)}{4}} \] \[\Large d = \sqrt{\frac{12a^2}{4}}\] \[\Large d = \sqrt{3a^2} \] \[\Large d = a\sqrt{3} \] Now do the same process for RB for R ( a/2,b/2) and B (2a,0)

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