The definition and properties of the perpendicular bisector are presented along with problems and their detailed solutions. Note: Figures are not drawn to scale.
Definition and Construction of Perpendicular Bisector
The perpendicular bisector of a line segment \(AB\) is a line that is perpendicular to \(AB\) and passes through the midpoint of segment \(AB\).
\[ \overline{MA} = \overline{MB} \]
To construct a perpendicular bisector of segment \(AB\):
- Draw two circles centered at \(A\) and \(B\) with equal radii larger than half the length of segment \(AB\).
- Connect the intersection points \(P\) and \(Q\) of the two circles to form a line. This line is perpendicular to \(AB\) and passes through the midpoint \(M\) of segment \(AB\).
Perpendicular Bisector of a Triangle and Circumcenter
Let \(ABC\) be a triangle where \(OM_{AB}\), \(OM_{BC}\), and \(OM_{AC}\) are the perpendicular bisectors of \(AB\), \(BC\), and \(AC\) respectively. The three bisectors intersect at the same point \(O\), called the circumcenter, which is the center of the circumcircle passing through vertices \(A\), \(B\), and \(C\).
The radius of the circumcircle \(R\) is given by: \[ R = \dfrac{1}{2} \dfrac{a}{\sin(A)} = \dfrac{1}{2} \dfrac{b}{\sin(B)} = \dfrac{1}{2} \dfrac{c}{\sin(C)} \]
Problems with Solutions
Problem 1: Angle and Area with Perpendicular Bisector
In the figure below, \(DE\) is the perpendicular bisector to segment \(AB\) whose length is equal to \(4\) units.
a) Find the size of angle \(x\).
b) Find the area of triangle \(ABC\).
Solution to Problem 1:
a) Since \(DE\) is the perpendicular bisector to segment \(AB\), \(DE\) and \(AB\) are perpendicular, so \(\angle CFA = 90^\circ\). The sum of angles in triangle \(AFC\) is \(180^\circ\):
Angles \(\angle ACF\) and \(x\) are vertical angles and therefore congruent; hence \(x = 54^\circ\).
b) \(FC\) is the height of triangle \(ABC\) and its length is given by:
\[ \overline{FC} = \overline{FA} \tan(36^\circ) \]Since \(DE\) is a perpendicular bisector and \(\overline{AB} = 4\):
\[ \overline{FA} = \dfrac{1}{2} \overline{AB} = 2 \]The area \(A\) of triangle \(ABC\) is:
\[ A = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times \overline{AB} \times \overline{FC} \] \[ A = \dfrac{1}{2} \times 4 \times 2 \tan(36^\circ) \approx 2.9 \text{ square units} \]Problem 2: Area of Quadrilateral Formed by Perpendicular Bisectors
In the figure below, \(DM\) is the perpendicular bisector to segment \(AB\) (length \(4\) units) and \(DN\) is the perpendicular bisector to segment \(BC\) (length \(6\) units). Find the area of the quadrilateral \(MDNB\) if segment \(BD = 7\) units.
Solution to Problem 2:
Calculate the area of quadrilateral \(MDNB\) as the sum of the areas of right triangles \(BDM\) and \(BDN\).
Since \(DM\) is a perpendicular bisector for \(AB\):
\[ \overline{MB} = \dfrac{1}{2} \overline{AB} = 2 \]Using the Pythagorean theorem in right triangle \(BDM\):
\[ \overline{BD}^2 = \overline{MB}^2 + \overline{MD}^2 \] \[ 7^2 = 2^2 + \overline{MD}^2 \implies \overline{MD} = \sqrt{45} = 3\sqrt{5} \]
Since \(DN\) is a perpendicular bisector for \(BC\):
\[ \overline{NB} = \dfrac{1}{2} \overline{BC} = 3 \]Using the Pythagorean theorem in right triangle \(BDN\):
\[ \overline{BD}^2 = \overline{NB}^2 + \overline{ND}^2 \] \[ 7^2 = 3^2 + \overline{ND}^2 \implies \overline{ND} = \sqrt{40} = 2\sqrt{10} \]Area of triangle \(BDM = \dfrac{1}{2} \overline{MB} \cdot \overline{MD} = \dfrac{1}{2} \cdot 2 \cdot 3\sqrt{5} = 3\sqrt{5}\)
Area of triangle \(BDN = \dfrac{1}{2} \overline{NB} \cdot \overline{ND} = \dfrac{1}{2} \cdot 3 \cdot 2\sqrt{10} = 3\sqrt{10}\)
Total Area \(A = 3\sqrt{5} + 3\sqrt{10} = 3(\sqrt{5} + \sqrt{10}) \approx 16.2 \text{ square units}\).
Problem 3: Ratio of Areas involving Trapezoid and Triangle
In the figure below, \(AD\) is parallel to \(BC\) and \(FE\) is the perpendicular bisector to segment \(AD\) whose length is equal to half the length of segment \(BC\). Find the ratio of the area of triangle \(AEF\) given that the area of the trapezoid \(ABCD\) is 120 square units.
Solution to Problem 3:
Since \(FE\) is a perpendicular bisector for \(AD\), \(\overline{AF} = \dfrac{1}{2}\overline{AD}\) and \(AEF\) is a right triangle.
Area \(A_1\) of triangle \(AEF\): \(A_1 = \dfrac{1}{2} \times \overline{AF} \times \overline{EF}\)
Area \(A_2\) of trapezoid \(ABCD\): \(A_2 = \dfrac{1}{2} \times (\overline{BC} + \overline{AD}) \times \overline{EF}\)
The ratio \(\dfrac{A_2}{A_1}\) simplifies to:
\[ \dfrac{A_2}{A_1} = \dfrac{\overline{BC} + \overline{AD}}{\overline{AF}} \]Given \(\overline{AD} = \dfrac{1}{2}\overline{BC}\), we have \(\overline{AF} = \dfrac{1}{2}\overline{AD} = \dfrac{1}{4}\overline{BC}\). Substituting this:
\[ \dfrac{A_2}{A_1} = \dfrac{\overline{BC} + \dfrac{1}{2}\overline{BC}}{\dfrac{1}{4}\overline{BC}} = \dfrac{1 + \dfrac{1}{2}}{\dfrac{1}{4}} = 6 \]Hence, \(A_1 = \dfrac{A_2}{6} = \dfrac{120}{6} = 20 \text{ square units}\).