Question

Given triangle ABC. Prove the three medians of triangle ABC intersect at a common point.

Answer 1

Answer:

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 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

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

IV Line 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.

III Line 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

I BGCH is a parallelogram -------- Properties of a Parallelogram (opposite sides are parallel)

Now we draw conclusions from that.

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

Answer: IV III I II, second choice