BIGDOG96
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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BIGDOG96
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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
BIGDOG96
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IV, II, III, I
II, IV, I, III
IV, II, I, III
II, IV, III, I
BIGDOG96
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It won't let me post the statements and justification in a two column
ganeshie8
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try taking a screenshot and post the pix... here you're trying to use parallelogram properties i think
BIGDOG96
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I just did.
BIGDOG96
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Does this work?
ganeshie8
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look at \(\triangle ABH\)
BIGDOG96
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Just heads up I HAVE NO CLUE what I'm doing.
ganeshie8
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FG is the midsegment , so it is parallel to BH
FG || BH
BIGDOG96
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ok
ganeshie8
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its okay im sure you would get enough clue by the time we finish ;p
BIGDOG96
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Alright, continue.
ganeshie8
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do the same thing for another triangle,
look at \(\triangle\) AHC
whats the midsegment in that triangle
BIGDOG96
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EB
BIGDOG96
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?
BIGDOG96
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No EG?
ganeshie8
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|dw:1347462588707:dw|
ganeshie8
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|dw:1347462742317:dw|
ganeshie8
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thats right, EG is the midsegment
ganeshie8
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so EG || HC
BIGDOG96
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Yay
ganeshie8
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good work :) so far.
so our first statement must be which one ?
BIGDOG96
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2?
ganeshie8
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Perfect !
ganeshie8
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lets continue our logic from this point
BIGDOG96
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There's only two that start with two.
ganeshie8
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only one of them is true
BIGDOG96
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And they both have four next
ganeshie8
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we have this so far :
we got two parallel sides according to midsegment theorem :
FG || BH
GE || HC
ganeshie8
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|dw:1347463059283:dw|
ganeshie8
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since GE || HC,
BG also parallel to HC
BG || HC
so next statement is IV as you said
ganeshie8
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what could be the next ?
BIGDOG96
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1?
ganeshie8
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|dw:1347463185726:dw|
ganeshie8
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thats one set of parallel sides
BIGDOG96
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Oh so 3?
ganeshie8
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|dw:1347463228999:dw|
ganeshie8
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thats another set of parallel sides
ganeshie8
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doesnt it make that piece a parallelogram ?
BIGDOG96
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Yes
BIGDOG96
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So I was right, it's 1?
ganeshie8
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yes ofcourse you're right :)
BIGDOG96
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Thanks
ganeshie8
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np.. yw ! (:
BIGDOG96
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D
BIGDOG96
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See you later. Thanks again!
ganeshie8
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is that a smiley D or you're saying the correct sequence is D lol
ganeshie8
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the correct sequence of statements is this :
II, IV, I, III
BIGDOG96
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BOTH
ganeshie8
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hmm
BIGDOG96
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LOL
ganeshie8
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good luck wid ur geometry proofs looks like you're at end of course. wish u best grades :) cya
BIGDOG96
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YEP thanks