Chapter 6.2, Problem 12PSB

To determine

To prove: The line BD is the perpendicular bisector of AC.

Explanation of Solution

Given information:

Line $\overleftrightarrow{\text{PB}}$ is perpendicular to a given plane (m)

Any point (D) is the midpoint of given line segment $\left(\overline{\text{AC}}\right)$

Triangle PAC is isosceles with base $\overline{\text{AC}}$

Formula used:

Congruency means same shape and same size

Congruent Part of Congruent Triangles are Congruent

Reflexive property

If a line is perpendicular to plane than all line passing through its foot will also perpendicular to it

Perpendicular lines always form right angles

Bisector divides the line segment in two equal parts

Proof:

  $\overleftrightarrow{\text{PB}}\perp \text{m}$ …… given

  $\overline{\text{PB}}\perp \overline{\text{BA}}\text{ \& }\overline{\text{PB}}\perp \overline{\text{BC}}$ ……. If a line is perpendicular to plane than all line passing through its foot will also perpendicular to it

  $\angle \text{PBC}=\angle \text{PAB}=90°$ ….. Perpendicular lines from right angles

  $\angle \text{PBC}\cong \angle \text{PAB}$ …… right angles are congruent, by definition

D is the midpoint of $\overline{\text{AC}}$ and $\Delta \text{PAC}$ is an isosceles triangle with base $\overline{\text{AC}}$ …. Given

  $\overline{\text{BC}}=\overline{\text{BD}}$ ……… If a point on the perpendicular bisector, it is equidistant from the segment’s end points

  $\therefore \overline{\text{BC}}\cong \overline{\text{BD}}$ …….by definition

  $\overline{\text{PB}}\cong \overline{\text{PB}}$ ……by reflexive property

  $\therefore \Delta \text{PBC}\cong \Delta \text{PBA}$ ……SAS property

  $\overleftrightarrow{\text{BD}}$ perpendicular bisector of $\overline{\text{AC}}$ ….. congruent parts of congruent triangles are congruent

Hence proved