Question

The Triangle Midsegment Theorem states that in a triangle, the segment joining the midpoints of any two sides will be parallel to the third side and half its length.
Provide the missing reasons for the proof of the second part of the triangle midsegment theorem, that the segment joining the midpoints of any two sides will be half the length of the third side.

Given: P is the midpoint of AB

Q is the midpoint of AC

Prove: BC=2PQ

Answer 1

Answer:

If the midpoints of any two sides of a triangle are joined by a segment, then that segment is parallel to the third side and half its length.

Step-by-step explanation:

We know the following equalities by the midpoint construction:

AD = DC and AE = EB. We can use the reflexive property of equality for angle A to obtain a SAS similarity of triangles DAE and CAB. Now that the two triangles are congruent, we have m<ADE = m<ACB. By the converse of the Corresponding Angles Theorem, we have DE || CB.

Now we need to show that DE = ½ CB.

Claim: Quadrilateral CBEF is a parallelogram. If we can show FC=EB, we can conclude that CBEF is a parallelogram.

*(side): We have m<FCD = m<EAD by the Alternate Interior Angles Theorem

*(angle): We have AD = DC (from above – definition of midpoint)

*(side): We have m<FDC = m<ADE by the Vertical Angles Theorem.

**We now have Triangle FDC is congruent to Triangle EDA by the SAS Congruence Theorem.

Therefore, FC = AE --> FC = EB. If a pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is parallelogram (previous theorem).  Therefore FE = CB. Also, the above triangle congruence gives us FD = DE.

Write: FE = FD + DE --> FE = 2DE --> DE = ½ FE

Since FE = CB, we can substitute: DE = ½ CB.

Therefore the Midsegment is half of the measure of the third side.

Conclusion: If the midpoints of any two sides of a triangle are joined by a segment, then that segment is parallel to the third side and half its length.