Question

4 which construction could be used to construct an isosceles triangle abc given line segment ab? a mark a third point c that is not on segment ab. draw segments ac and bc. b label a point c on segment ab and construct a line perpendicular to segment ab through point c. draw segments ac and bc. c construct the perpendicular bisector of segment ab. mark the intersection of this line and segment ab and label it c. draw segments ac and bc. d construct the perpendicular bisector of segment ab. mark any point c on the perpendicular bisector except where it intersects segment ab. draw segments ac and bc.

Explanation:

Step1: Recall isosceles - triangle property

An isosceles triangle has two equal - length sides. For a triangle $\triangle ABC$ with base $AB$, we want $AC = BC$.

Step2: Recall perpendicular - bisector property

The perpendicular bisector of a line segment is the set of all points that are equidistant from the two endpoints of the line segment. If point $C$ lies on the perpendicular bisector of segment $AB$, then by the definition of the perpendicular bisector, $AC=BC$.

Step3: Analyze each option

• Option A: Just marking a random point $C$ not on $AB$ does not guarantee $AC = BC$.
• Option B: Labeling a point $C$ on $AB$ and constructing a perpendicular to $AB$ through $C$ will not result in an isosceles triangle with $AB$ as the base.
• Option C: Marking the intersection of the perpendicular bisector and $AB$ will result in a degenerate triangle (a single - line segment), not a non - degenerate isosceles triangle.
• Option D: Marking any point $C$ on the perpendicular bisector of segment $AB$ (except the intersection with $AB$) and drawing segments $AC$ and $BC$ will result in a non - degenerate isosceles triangle since $AC = BC$ due to the property of the perpendicular bisector.

Answer:

D. Construct the perpendicular bisector of segment $AB$. Mark any point $C$ on the perpendicular bisector except where it intersects segment $AB$. Draw segments $AC$ and $BC$.