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All courses Math I · G-CO.12 35 of 59
Perform formal constructions with compass/straightedge, paper folding, string, reflective devices, or geometry software.

Find a midpoint using construction reasoning

Problem
Use the displayed perpendicular-bisector construction to identify the midpoint of AB and justify it from the point's incidence and equal distances to A and B.
Construction arcs intersect at U and V; line UV intersects segment AB at labeled point M. Open full size
Your answer
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Hint

Use point \(M\), where the perpendicular bisector meets segment \(AB\).

The perpendicular bisector of a segment crosses the segment at its midpoint, because every point on the perpendicular bisector is equidistant from the two endpoints.

Solution walkthrough

01

Inspect the construction incidence

\[M~\text{lies}~\text{on}~\text{segment}~AB~\text{and}~\text{on}~\text{line}~UV\]

The actual diagram places M at the intersection of horizontal segment AB and the constructed vertical line through arc intersections U and V.

02

Use equal-radius construction evidence

\[UA=UB~\text{and}~VA=VB\]

The paired arcs are drawn with equal radii from A and B, so both arc-intersection points are equidistant from the segment endpoints. Line UV is therefore the perpendicular bisector of AB.

03

Apply the perpendicular-bisector property

\[UV~\text{meets}~AB~\text{at}~M~->~MA=MB\]

A perpendicular bisector crosses its segment at the point equidistant from both endpoints, so M divides AB into equal lengths.

04

State the midpoint and justification

\[\text{Point}~M~\text{is}~\text{the}~\text{midpoint}~\text{of}~AB~\text{because}~\text{it}~\text{is}~\text{where}~\text{the}~\text{perpendicular}~\text{bisector}~\text{meets}~AB,~\text{so}~MA~=~MB.\]

The conclusion names M, its incidence on AB, and the equality that proves midpoint status.

+

Another way

  1. Use symmetry of the equal-radius arcs about their common chord UV to show it bisects AB at M.

!

Common mistake

Do not call U or V the midpoint. They are arc intersections off segment AB; the midpoint must lie on AB at M.

Completed construction: equal-radius arcs from A and B meet at U and V; line UV crosses segment AB at M; because M lies on the perpendicular bisector, MA=MB, so M is the midpoint of AB.
midpoint: M; reason: M is where the perpendicular bisector meets AB, so M is equidistant from A and B