High School

In triangle ABC, show that the midsegment ED is parallel to BC and that ED = \(\frac{1}{2}\)BC.

A. ED || BC and ED = \(\frac{1}{2}\)BC
B. ED || AC and ED = \(\frac{1}{2}\)AC
C. ED || AB and ED = \(\frac{1}{2}\)AB
D. ED || BC and ED = BC

Answer :

To show that the midsegment ED is parallel to BC and that ED = 1/2BC in triangle ABC, we can use the properties of midsegments.

To show that the midsegment ED is parallel to BC and that ED = 1/2BC in triangle ABC, we can use the properties of midsegments.

Firstly, a midsegment of a triangle is a segment connecting the midpoints of two sides. In triangle ABC, let E be the midpoint of AB and D be the midpoint of AC.

Since E is the midpoint of AB and D is the midpoint of AC, we can conclude that DE is parallel to BC and DE = 1/2BC. Therefore, the correct option is a. ED || BC and ED = 1/2BC.

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