Use ∆ABC to answer the question that follows.
Given: ∆ABC
Prove: The three medians of ∆ABC intersect at a common point.
When written in the correct order, the two-column proof below describes the statements and justifications for proving the three medians of a triangle all intersect in one point.
Which is the most logical order of statements and justifications I, II, III, and IV to complete the proof?

- anonymous

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- anonymous

Statements
Justifications
Point F is a midpoint of Line segment AB
Point E is a midpoint of Line segment AC
Draw Line segment BE
Draw Line segment FC
by Construction
Point G is the point of intersection between Line segment BE and Line segment FC
Intersecting Lines Postulate
Draw Line segment AG
by Construction
Point D is the point of intersection between Line segment AG and Line segment BC
Intersecting Lines Postulate
Point H lies on Line segment AG such that Line segment AG ≅ Line segment GH
by Construction
I
BGCH is a parallelogram
Properties of a Parallelogram (opposite sides are parallel)
II
Line segment FG is parallel to line segment BH and Line segment GE is parallel to line segment HC
Midsegment Theorem
III
Line segment BD ≅ Line segment DC
Properties of a Parallelogram (diagonals bisect each other)
IV
Line segment GC is parallel to line segment BH and Line segment BG is parallel to line segment HC
Substitution
Line segment AD is a median
Definition of a Median

##### 1 Attachment

- anonymous

IV, II, III, I
II, IV, I, III
IV, II, I, III
II, IV, III, I

- anonymous

It won't let me post the statements and justification in a two column

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## More answers

- ganeshie8

try taking a screenshot and post the pix... here you're trying to use parallelogram properties i think

- anonymous

I just did.

- anonymous

Does this work?

##### 1 Attachment

- ganeshie8

look at \(\triangle ABH\)

- anonymous

Just heads up I HAVE NO CLUE what I'm doing.

- ganeshie8

FG is the midsegment , so it is parallel to BH
FG || BH

- anonymous

ok

- ganeshie8

its okay im sure you would get enough clue by the time we finish ;p

- anonymous

Alright, continue.

- ganeshie8

do the same thing for another triangle,
look at \(\triangle\) AHC
whats the midsegment in that triangle

- anonymous

EB

- anonymous

?

- anonymous

No EG?

- ganeshie8

|dw:1347462588707:dw|

- ganeshie8

|dw:1347462742317:dw|

- ganeshie8

thats right, EG is the midsegment

- ganeshie8

so EG || HC

- anonymous

Yay

- ganeshie8

good work :) so far.
so our first statement must be which one ?

- anonymous

2?

- ganeshie8

Perfect !

- ganeshie8

lets continue our logic from this point

- anonymous

There's only two that start with two.

- ganeshie8

only one of them is true

- anonymous

And they both have four next

- ganeshie8

we have this so far :
we got two parallel sides according to midsegment theorem :
FG || BH
GE || HC

- ganeshie8

|dw:1347463059283:dw|

- ganeshie8

since GE || HC,
BG also parallel to HC
BG || HC
so next statement is IV as you said

- ganeshie8

what could be the next ?

- anonymous

1?

- ganeshie8

|dw:1347463185726:dw|

- ganeshie8

thats one set of parallel sides

- anonymous

Oh so 3?

- ganeshie8

|dw:1347463228999:dw|

- ganeshie8

thats another set of parallel sides

- ganeshie8

doesnt it make that piece a parallelogram ?

- anonymous

Yes

- anonymous

So I was right, it's 1?

- ganeshie8

yes ofcourse you're right :)

- anonymous

Thanks

- ganeshie8

np.. yw ! (:

- anonymous

D

- anonymous

See you later. Thanks again!

- ganeshie8

is that a smiley D or you're saying the correct sequence is D lol

- ganeshie8

the correct sequence of statements is this :
II, IV, I, III

- anonymous

BOTH

- ganeshie8

hmm

- anonymous

LOL

- ganeshie8

good luck wid ur geometry proofs looks like you're at end of course. wish u best grades :) cya

- anonymous

YEP thanks

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