Question Video: Completing a Proof Using the Triangle Midsegment Theorem Mathematics • First Year of Preparatory School

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.