Question

Complete the following proof.

Prove: In an isosceles triangle two medians are equal.

https://wca.sooschools.com/media/g_geo_ccss_2016/9/q413a.gif
https://wca.sooschools.com/media/g_geo_ccss_2016/9/q413bnew.gif

Answer 1

@563blackghost Can you help with this???

Answer 2

Or do u know someone who can <_< lol..

Answer 3

I can help, it's just trying to organize it seems like a lot O.O

Answer 4

Yeah I know .-. I dont see how assignments like this are legal :'D

Answer 5

x'D

Answer 6

First we need to identify what mid-point formula is:

\(\Large\bf{(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}-y_{2}}{2})}\)

We first find the midpoint of AB. Point A is `(0,0)` and Point B is `(2a,0)`. Can you plug this and simplify?

Answer 7

Oh wait! I'm sorry I forgot to add. I have down to N done and its correct. Its just M, MA, NB that are incorrect and I cant figure out what they should be. I shouldve added that :x

Answer 8

So what did you get for N?

Answer 9

http://prntscr.com/hnm5ts

Answer 10

why is math so complicated

Answer 11

Right... <_<

Answer 12

Like I'm good at every other subject, besides math :')

Answer 13

in advanced math they take 1+1=2 to 1+1=3

Answer 14

like wth

Answer 15

Haha.. Omg. I'm already about to loose it with math. much less if they start doing that XD

Answer 16

To find M as the midpoint you need to find the midpoint of CB.

Point C is `(a,b)` and Point B is (2a,0).

\(\Large\bf{\frac{2a+a}{2}, \frac{b+0}{2}}\)

Answer 17

why must you explain well x-x

Answer 18

So the midpoint would be 3a/2, b/2?

Answer 19

yes because on the graph point M is shown to be 3a/2, b/2

Answer 20

Now I do believe we have to find the distance (the format of equation MA is really similar to it).

We identify that Point A is `(0,0)` amd Point M is \(\large\bf{(\frac{3a}{2}, \frac{b}{2})}\). Plug this in distance formula...

\(\Large\bf{MA=\sqrt{(\frac{3a}{2}-b)^{2}+({\frac{b}{2}+0)^{2}}}}\)

Answer 21

WAIT THAT EQUATION IS WRONG

Answer 22

\(\Large\bf{MA=\sqrt{(\frac{3a}{2}-0)^{2}+({\frac{b}{2}-0)^{2}}}}\)

Answer 23

....wait no im confused on the next equation >.<

Answer 24

Wait is it wrong or right? I'm confuzzled now too :') okay I stay confused with math xD

Answer 25

I'm pretty sure that equation is right for MA.

Answer 26

OHHHH i didnt realize there is an equal sign :P

Yea the equation for MA is correct.

Answer 27

This would simplify to be...

\(\Large\bf{\sqrt{(\frac{3a}{2})^{2} + (\frac{b}{2})^{2}}}\) in which we get...

\(\Large\bf{\sqrt{\frac{9a^{2}}{4} + \frac{b^{2}}{4}}}\)

Answer 28

You follow the same process for NB.

Distance Formula.

\(\Large\bf{\sqrt{(\frac{a}{2}-2a)^{2}+({\frac{b}{2}-0)^{2}}}}\)

Now we simplify....

\(\Large\bf{\sqrt{(\frac{a}{2}-\frac{2a}{2})^{2}+({\frac{b}{2}-0)^{2}}}}\)

\(\Large\bf{\sqrt{(\frac{-3a}{2})^{2}+({\frac{b}{2})^{2}}}}\)

\(\Large\bf{\sqrt{\frac{9a^{2}}{4}+{\frac{b^{2}}{4}}}}\)

Answer 29

THERE DONE!

Answer 30

XD

Answer 31

omg thank you so much you dont know how much that just helped me :')

Answer 32

your welcome XD