Video Transcript
In the given figure, 𝐸 and 𝐹 are
the midpoints of line segments 𝐴𝐵 and 𝐴𝐶, respectively, 𝐵𝐷 equals one-half
𝐵𝐶, and 𝐵 lies on line segment 𝐷𝐶. What is the shape of 𝐸𝐹𝐵𝐷?
We can begin this question by
noting that the diagram shows the information that we have line segments that are
divided into two congruent pieces, since 𝐸 is the midpoint of line segment 𝐴𝐵 and
𝐹 is the midpoint of line segment 𝐴𝐶. The fact that we have these two
midpoints might indicate that one of the triangle midsegment theorems can be
applied: this one in particular that the line segment joining the midpoints of two
sides of a triangle is parallel to the third side and is half its length.
So, if we consider triangle 𝐴𝐵𝐶,
line segment 𝐸𝐹 is a line segment connecting the midpoints of two sides of a
triangle. Therefore, it is parallel to the
third side, which is 𝐵𝐶, and 𝐸𝐹 must be half the length of this side 𝐵𝐶. And notice that we are given
another side which is also equal to one-half 𝐵𝐶: this line segment, 𝐵𝐷.
So now, if we consider the
quadrilateral 𝐸𝐹𝐵𝐷, we know that this quadrilateral has a pair of congruent
sides. And we can also say that these two
sides are parallel, because we determined that line segment 𝐸𝐹 was parallel to
line segment 𝐵𝐶. And we know that 𝐵 lies on the
line segment 𝐷𝐶.
Now, the shape 𝐸𝐹𝐵𝐷 does look
like a parallelogram, and in fact what we have shown here would prove this, because
one way we can prove a quadrilateral is a parallelogram is by showing that one pair
of opposite sides in a quadrilateral are both parallel and congruent. Therefore, we can give the answer
that 𝐸𝐹𝐵𝐷 is a parallelogram.
