Chapter 6.2, Problem 10PSB

To determine

To prove: The given line BC bisects the segment AC

Explanation of Solution

Given information:

Given line segments are perpendicular to a given plane, i.e., $\overline{AB}\perp \text{n}$ and $\overline{\text{CD}}\perp \text{n}$

Given line bisects the given line segment, i.e., $\overleftrightarrow{\text{AC}}\text{ bisects }\overline{\text{BD}}$

Formula used:

Congruency means same shape and same size

Congruent Parts 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:

  $\overline{AB}\perp \text{n}$ and $\overline{\text{CD}}\perp \text{n}$ ........... given

  $\overleftrightarrow{\text{AC}}\text{ bisects }\overline{\text{BD}}$ ........... given

  $\angle \text{ABE}=\angle \text{CDE}=90°$ ........... perpendicular lines form right angle

  $\overline{\text{AB}}\cong \overline{\text{CD}}$ ........... given

  $\overline{\text{BE}}\cong \overline{\text{ED}}$ ........... reflexive property

  $\therefore$$\Delta \text{ABE}\cong \Delta \text{CDE}$ and

  $\therefore$$\overleftrightarrow{\text{BD}}\text{ bisects }\overline{\text{AC}}$

Hence proved