The proof for the Trapezoid Midsegment Theorem is partially filled in. Supply the missing reason to prove that the length of the midsegment is half the sum of the lengths of the bases.(1 point) ST¯¯¯¯¯¯¯ is the midsegment of trapezoid WXYZ. given SU¯¯¯¯¯¯¯∥WX¯¯¯¯¯¯¯¯¯¯ UT¯¯¯¯¯¯¯∥ZY¯¯¯¯¯¯¯definition of midsegment (trapezoid) SU¯¯¯¯¯¯¯ is the midsegment of △WXZ.UT¯¯¯¯¯¯¯ is the midsegment of △XYZ.definition of midsegment (triangle)SU=12WXUT=12ZYMidsegment TheoremST=SU+UT ST=12WX+12ZY substitutionST=12(WX+ZY) Distributive Property

Answer 1

The missing reason for the step "ST = SU + UT" is the Segment Addition Postulate. This postulate states that the length of a segment is equal to the sum of the lengths of its parts.

So, the completed proof with the missing reason would include:

1. ST = SU + UT (Segment Addition Postulate)

Here's the complete argument for the theorem:

• Given: SU is the midsegment of trapezoid WXYZ.
• SU || WX and UT || ZY (definition of midsegment in trapezoid).
• SU is the midsegment of triangle WXZ.
• UT is the midsegment of triangle XYZ (definition of midsegment in triangle).
• SU = 1/2 WX and UT = 1/2 ZY (Midsegment Theorem).
• ST = SU + UT (Segment Addition Postulate).
• ST = 1/2 WX + 1/2 ZY (substitution).
• ST = 1/2 (WX + ZY) (Distributive Property).

This shows that the length of the midsegment ST is half the sum of the lengths of the bases WX and ZY.