Use δabc to answer the question that follows: given: δabc prove: the three medians of δabc 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: 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 bd ≅ line segment dc properties of a parallelogram (diagonals bisect each other) iii line segment gc is parallel to line segment bh and line segment bg is parallel to line segment hc substitution iv line segment fg is parallel to line segment bh and line segment ge is parallel to line segment hc midsegment theorem line segment ad is a median definition of a median which is the most logical order of statements and justifications i, ii, iii, and iv to complete the proof? iii, iv, ii, i iv, iii, i, ii iii, iv, i, ii iv, iii, ii, i

Let's try to render the first part of the proof a bit more legibly.

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 FCby Construction

Point G is the point of intersection between Line segment BE and Line segment FCIntersecting Lines Postulate

Draw Line segment AGby Construction

Point D is the point of intersection between Line segment AG and Line segment BCIntersecting Lines Postulate

Point H lies on Line segment AG such that Line segment AG ≅ Line segment GHby Construction

OK, now we continue. We need to prove some parallel lines; statement 4 lets us do so.

IVLine segment FG is parallel to line segment BH and Line segment GE is parallel to line segment HC Midsegment Theorem

Now that we've shown some segments parallel we extend that to collinear segments.

IIILine segment GC is parallel to line segment BH and Line segment BG is parallel to line segment HC Substitution

We have enough parallel lines to prove a parallelogram

IBGCH is a parallelogram Properties of a Parallelogram (opposite sides are parallel)

Now we draw conclusions from that.

IILine segment BD ≅ Line segment DC Properties of a Parallelogram (diagonals bisect each other)

IV III I II, second choice

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.

step-by-step explanation: