Here's the question you clicked on:

## BIGDOG96 Group Title 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? 2 years ago 2 years ago

• This Question is Closed
1. BIGDOG96

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

2. BIGDOG96

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

3. BIGDOG96

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

4. ganeshie8

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

5. BIGDOG96

I just did.

6. BIGDOG96

Does this work?

7. ganeshie8

look at \(\triangle ABH\)

8. BIGDOG96

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

9. ganeshie8

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

10. BIGDOG96

ok

11. ganeshie8

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

12. BIGDOG96

Alright, continue.

13. ganeshie8

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

14. BIGDOG96

EB

15. BIGDOG96

?

16. BIGDOG96

No EG?

17. ganeshie8

|dw:1347462588707:dw|

18. ganeshie8

|dw:1347462742317:dw|

19. ganeshie8

thats right, EG is the midsegment

20. ganeshie8

so EG || HC

21. BIGDOG96

Yay

22. ganeshie8

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

23. BIGDOG96

2?

24. ganeshie8

Perfect !

25. ganeshie8

lets continue our logic from this point

26. BIGDOG96

There's only two that start with two.

27. ganeshie8

only one of them is true

28. BIGDOG96

And they both have four next

29. ganeshie8

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

30. ganeshie8

|dw:1347463059283:dw|

31. ganeshie8

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

32. ganeshie8

what could be the next ?

33. BIGDOG96

1?

34. ganeshie8

|dw:1347463185726:dw|

35. ganeshie8

thats one set of parallel sides

36. BIGDOG96

Oh so 3?

37. ganeshie8

|dw:1347463228999:dw|

38. ganeshie8

thats another set of parallel sides

39. ganeshie8

doesnt it make that piece a parallelogram ?

40. BIGDOG96

Yes

41. BIGDOG96

So I was right, it's 1?

42. ganeshie8

yes ofcourse you're right :)

43. BIGDOG96

Thanks

44. ganeshie8

np.. yw ! (:

45. BIGDOG96

D

46. BIGDOG96

See you later. Thanks again!

47. ganeshie8

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

48. ganeshie8

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

49. BIGDOG96

BOTH

50. ganeshie8

hmm

51. BIGDOG96

LOL

52. ganeshie8

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

53. BIGDOG96

YEP thanks