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anonymous
 3 years ago
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 twocolumn 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
 3 years ago
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 twocolumn 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?

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Statements 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0IV, II, III, I II, IV, I, III IV, II, I, III II, IV, III, I

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It won't let me post the statements and justification in a two column

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6try taking a screenshot and post the pix... here you're trying to use parallelogram properties i think

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6look at \(\triangle ABH\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Just heads up I HAVE NO CLUE what I'm doing.

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6FG is the midsegment , so it is parallel to BH FG  BH

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6its okay im sure you would get enough clue by the time we finish ;p

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6do the same thing for another triangle, look at \(\triangle\) AHC whats the midsegment in that triangle

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6dw:1347462588707:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6dw:1347462742317:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6thats right, EG is the midsegment

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6good work :) so far. so our first statement must be which one ?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6lets continue our logic from this point

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There's only two that start with two.

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6only one of them is true

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And they both have four next

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6we have this so far : we got two parallel sides according to midsegment theorem : FG  BH GE  HC

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6dw:1347463059283:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6since GE  HC, BG also parallel to HC BG  HC so next statement is IV as you said

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6what could be the next ?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6dw:1347463185726:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6thats one set of parallel sides

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6dw:1347463228999:dw

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6thats another set of parallel sides

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6doesnt it make that piece a parallelogram ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So I was right, it's 1?

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6yes ofcourse you're right :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0See you later. Thanks again!

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6is that a smiley D or you're saying the correct sequence is D lol

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6the correct sequence of statements is this : II, IV, I, III

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.6good luck wid ur geometry proofs looks like you're at end of course. wish u best grades :) cya
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