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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?

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

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Other answers:

try taking a screenshot and post the pix... here you're trying to use parallelogram properties i think
I just did.
Does this work?
1 Attachment
look at \(\triangle ABH\)
Just heads up I HAVE NO CLUE what I'm doing.
FG is the midsegment , so it is parallel to BH FG || BH
its okay im sure you would get enough clue by the time we finish ;p
Alright, continue.
do the same thing for another triangle, look at \(\triangle\) AHC whats the midsegment in that triangle
No EG?
thats right, EG is the midsegment
so EG || HC
good work :) so far. so our first statement must be which one ?
Perfect !
lets continue our logic from this point
There's only two that start with two.
only one of them is true
And they both have four next
we have this so far : we got two parallel sides according to midsegment theorem : FG || BH GE || HC
since GE || HC, BG also parallel to HC BG || HC so next statement is IV as you said
what could be the next ?
thats one set of parallel sides
Oh so 3?
thats another set of parallel sides
doesnt it make that piece a parallelogram ?
So I was right, it's 1?
yes ofcourse you're right :)
np.. yw ! (:
See you later. Thanks again!
is that a smiley D or you're saying the correct sequence is D lol
the correct sequence of statements is this : II, IV, I, III
good luck wid ur geometry proofs looks like you're at end of course. wish u best grades :) cya
YEP thanks

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