Question

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 ab
exercise #5: the figure shown below is known as a kite in geometry (for obvious reasons). in kite mnpq, it is known that pm is the perpendicular bisector of nq
list the different pieces of information we can conclude from knowing that pm is the perpendicular bisector of nq
exercise #6: in the diagram shown below, it is known that agb, egf, cgd, ab = 28, ag = 14, m∠egd = 26°, and m∠bgf = 64°. is cd the perpendicular bisector of ab? justify your answer.
n - gen math geometry - unit 1 - beginning concepts - lesson 8
cmathstruction, rd hook, ny 11871, c2013

Explanation:

Exercise #4:

1. Find mid - point: Place the ruler along the line segment \(AB\). Measure the length of \(AB\). Divide the length by 2 to find the mid - point. Mark the mid - point.
2. Draw perpendicular bisector: Place the set - square or protractor at the mid - point such that one of its sides is along the line segment \(AB\). Draw a line perpendicular to \(AB\) through the mid - point. This is the perpendicular bisector of \(AB\).

Exercise #5:

1. Equal segments: Since \(PM\) is the perpendicular bisector of \(NQ\), we can conclude that \(NR = RQ\) (a perpendicular bisector divides a line segment into two equal parts).
2. Right angles: \(\angle MRN=\angle MRQ = 90^{\circ}\) (by the definition of a perpendicular bisector). Also, any point on the perpendicular bisector is equidistant from the endpoints of the line segment it bisects, so \(PN=PQ\) and \(MN = MQ\).

Exercise #6:

1. Check mid - point condition: Given \(AB = 28\) and \(AG=14\), \(G\) is the mid - point of \(AB\) since \(AG=\frac{AB}{2}\).
2. Check perpendicular condition: We know that \(\angle EGD = 20^{\circ}\) and \(\angle BGF=64^{\circ}\). Since \(\angle EGD\) and \(\angle BGF\) are vertical angles, they are equal. But we need to check if \(\angle AGC = 90^{\circ}\) or \(\angle BGC = 90^{\circ}\). There is no information given to suggest that \(\angle AGC\) or \(\angle BGC\) is \(90^{\circ}\). So, \(CD\) is not the perpendicular bisector of \(AB\) because we do not have evidence that it is perpendicular to \(AB\) even though it passes through the mid - point \(G\) of \(AB\).

Answer:

• Exercise #4: Follow the steps above to draw the perpendicular bisector of \(AB\).
• Exercise #5: \(NR = RQ\), \(\angle MRN=\angle MRQ = 90^{\circ}\), \(PN = PQ\), \(MN=MQ\).
• Exercise #6: \(CD\) is not the perpendicular bisector of \(AB\) because there is no evidence that it is perpendicular to \(AB\) although it passes through the mid - point of \(AB\).