Question

we can combine the idea of a segment bisector with that of perpendicular lines to introduce another important geometric concept, that of a perpendicular bisector
perpendicular bisector
a line, ray, or segment that both bisects a line segment and is perpendicular to the line segment is called a perpendicular bisector. every line segment has a unique line that is the segments perpendicular bisector.
exercise #4: the same segment from exercise #1 is shown below. use a ruler to again find its mid - point. then, use a protractor to help draw the line that is the perpendicular bisector of $overline{ab}$.
exercise #5: the figure shown below is known as a kite in geometry (for obvious reasons). in kite $mnpq$, it is known that $overline{pm}$ is the perpendicular bisector of $overline{nq}$.
list the different pieces of information we can conclude from knowing that $overline{pm}$ is the perpendicular bisector of $overline{nq}$.
exercise #6: in the diagram shown below, it is known that $overline{agb}$, $overline{egf}$, $overline{cgd}$, $ab = 28$, $ag = 14$, $mangle egd=26^{circ}$, and $mangle bgf = 64^{circ}$. is $overline{cd}$ the perpendicular bisector of $overline{ab}$? justify your answer.

Explanation:

Exercise #4

1. Find mid - point:

Place the ruler along segment $\overline{AB}$. Measure the length of $\overline{AB}$. Divide the length by 2 and mark the mid - point. Let the mid - point be $M$.

1. Draw perpendicular bisector:

Place the protractor at the mid - point $M$. Align the protractor's baseline with $\overline{AB}$. Mark a $90^{\circ}$ angle on either side of $\overline{AB}$ at $M$. Then draw a straight line through the marked $90^{\circ}$ points. This is the perpendicular bisector of $\overline{AB}$.

Exercise #5

1. Mid - point property:

Since $\overline{PM}$ is the perpendicular bisector of $\overline{NQ}$, we know that $NR = RQ$ (a perpendicular bisector bisects the line segment).

1. Perpendicular property:

$\angle MRN=\angle MRQ = 90^{\circ}$ (a perpendicular bisector is perpendicular to the line segment).

1. Congruent triangles and other properties:

Triangles $\triangle MRN$ and $\triangle MRQ$ are right - angled triangles. Also, any point on $\overline{PM}$ is equidistant from $N$ and $Q$. For example, if we take a point $P$ on $\overline{PM}$, then $PN=PQ$.

Exercise #6

1. Check bisection:

Given $AB = 28$ and $AG=14$, then $GB=AB - AG=28 - 14 = 14$. So $AG = GB$, which means $\overline{CD}$ bisects $\overline{AB}$.

1. Check perpendicularity:

We know that $\angle EGD=26^{\circ}$ and $\angle BGF = 64^{\circ}$. Since $\angle EGD$ and $\angle BGF$ are vertical angles, and $\angle EGD+\angle DGB = 180^{\circ}-\angle BGF$. Also, $\angle DGB = 180^{\circ}-\angle BGF-\angle EGD=180^{\circ}-64^{\circ}-26^{\circ}=90^{\circ}$. So $\overline{CD}\perp\overline{AB}$.
Since $\overline{CD}$ bisects $\overline{AB}$ and $\overline{CD}\perp\overline{AB}$, $\overline{CD}$ is the perpendicular bisector of $\overline{AB}$.

Answer:

For Exercise #4: Follow the steps above to find mid - point and draw perpendicular bisector.
For Exercise #5: $NR = RQ$, $\angle MRN=\angle MRQ = 90^{\circ}$, and any point on $\overline{PM}$ is equidistant from $N$ and $Q$ (e.g., $PN = PQ$).
For Exercise #6: $\overline{CD}$ is the perpendicular bisector of $\overline{AB}$ because it bisects $\overline{AB}$ ($AG = GB$) and is perpendicular to $\overline{AB}$ ($\angle DGB = 90^{\circ}$).