Question

third student challenge: the proof is in the distance
step 1: create a line segment and label it $overline{ab}$.
step 2: create a perpendicular bisector for the line segment and label it $overline{cd}$.
step 3: create two additional points collinear with $overline{cd}$ and label them point e and point f respectively.
prove point e and point f are equidistant from point a and point b.
be prepared to share your ideas with the class.

Explanation:

Step1: Recall the property of perpendicular bisector

A perpendicular bisector of a line - segment $\overline{AB}$ is a line that is perpendicular to $\overline{AB}$ and divides $\overline{AB}$ into two equal parts. Let the mid - point of $\overline{AB}$ be $M$ where $\overline{CD}$ intersects $\overline{AB}$. So, $AM = BM$ and $\angle CMA=\angle CMB = 90^{\circ}$.

Step2: Consider right - triangles

For any point $P$ on the perpendicular bisector $\overline{CD}$, we can form right - triangles $\triangle PMA$ and $\triangle PMB$. In right - triangles $\triangle PMA$ and $\triangle PMB$, we have $AM = BM$ (by the definition of the perpendicular bisector), $PM$ is common to both triangles, and $\angle PMA=\angle PMB = 90^{\circ}$.

Step3: Apply the Pythagorean theorem

By the Pythagorean theorem, in $\triangle PMA$, $PA^{2}=PM^{2}+AM^{2}$, and in $\triangle PMB$, $PB^{2}=PM^{2}+BM^{2}$. Since $AM = BM$, we have $PA^{2}=PB^{2}$, and thus $PA = PB$.

Step4: Substitute points E and F

Since points $E$ and $F$ are on the perpendicular bisector $\overline{CD}$ (they are collinear with $\overline{CD}$), for point $E$, $EA = EB$ and for point $F$, $FA = FB$.

Answer:

Points $E$ and $F$ are equidistant from points $A$ and $B$ because any point on the perpendicular bisector of a line - segment is equidistant from the endpoints of the line - segment.