Geometry: Perpendicular and Angle Bisector Theorems

PERPENDICULAR BISECTOR THEOREM

Theorems

• Perpendicular Bisector Theorem: If a point lies on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

  • If $CD \perp AB$ and $AD = BD$, then $CA = CB$ (where $CA=CB$ indicates points C, A, and B are equidistant).

• Converse of the Perpendicular Bisector Theorem: If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

  • If $CA = CB$, then a line exists through C such that $CD \perp AB$ and $AD = BD$ (where $AD=BD$).

Example Problems

1. Find the value of x.

   • Given equation: $7x + 24 = 13x - 48$

   • Solution:

     • Rearranging gives: $72 = 6x$

     • Therefore, $x = 12$.

2. Second Equational Example:

   • Problem: $17x - 3 = 9x + 13$

   • Simplifying: $8x = 16$

   • Thus, $x = 2$.

3. Find RS.

   • Given: $7x - 22 = 4x + 5$

   • Solution:

     • Combining terms yields: $3x = 27$

     • Hence, $x = 9$.

     • Final calculation: $RS = 4(9) + 5 = 41$.

4. Find AB.

   • Problem: $Sx - 11 = 3x + 17$

   • Solution involves:

     • $2x = 28 \ x = 14$.

5. Find EG.

   • Equation: $2x + 11 = 14x - 37$

   • Steps:

     • Rearranging yields $48 = 12x$ hence $x = 4$.

     • Final Answer: $EG = 2(4) + 11 = 19$.

6. Find JK.

   • Equations to solve:

     • From $5x - 11 = 3x + 17$,

     • deducing gives $6x - 26 = 3x + 34$ showing: $AB = 5(14) - 11 + 3(14) + 17 = 118$.

PERPENDICULAR BISECTOR PROBLEM EXAMPLE

7. Determine if point L(-9, 2) lies on the perpendicular bisector of line segment JK formed by points J(-7, -8) and K(1, 4).

   • Calculate midpoint JL:

     • $JL = (-7 + 9)^{2} + (-8 - 2)^{2} = \sqrt{4 + 100} = \sqrt{104}$.

     • Next, calculate distance KL: $KL = \sqrt{(1 + 9)^{2} + (4 - 2)^{2}} = \sqrt{100 + 4} = \sqrt{104}$. Confirming: Yes, by Perpendicular Bisector Converse.

ANGLE BISECTOR THEOREM

• Angle Bisector Theorem: If a point is on a bisector of an angle, then the point is equidistant from the sides of the angle.

  • If $AD$ bisects $\angle BAC$, then $AB = BD$ and $AC = CD$, concluding $BD = CD$ (where $BD=CD$ indicates balance in distances from the angle sides).

• Converse of the Angle Bisector Theorem: If a point is on the interior of an angle and equidistant from the sides of the angle, then it lies on the angle bisector.

  • If $BD = CD$ with $AB = BD$ and $AC = CD$, it follows that $\angle BAD = \angle CAD$.

Example Problems

8. Find the value of x.

   • Equation: $4x + 30 = 9x - 5$

   • Solving: $35 = 5x$ \ $x = 7$.

9. Another x problem:

   • $5x - 11 = 8x - 4$

   • Solve results in $30 = 3x \ x = 10$.

   • Identify $\angle BAD$ versus $\angle CAD$: $m\angle BAD = m\angle CAD$.

10. Find AD.

    • Given: $13x - 4 = 8x$ \ placing: $5x = 15 \ x = 3$, concluding $AD = 13(3) - 4 = 39 - 4$.

11. Finding angle values:

    • $m\angle XWZ, equation$:

      • For setup use: $8x - 32 = 3x + 3$ resulting in: $5x = 35\text{, which means } x = 7$.

12. Finding $m/FGH$ via the equation: $2x + 20 = 5x - 37$

    • Resulting in $7x = 19$, thus $m\angle FGH = 116$ degrees.