Chapter 6.2, Problem 4PSA

To determine

To prove: The given ray PV bisects the given angle TPS

Explanation of Solution

Given information:

Given line is perpendicular to given plane, i.e., $\overleftrightarrow{\text{TS}}\perp \text{m}$

Given line $\overline{\text{PV }}$ bisects $\overline{\text{TS}}$

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:

  $\overleftrightarrow{\text{TS}}\perp \text{m}$ ……. given

  $\overline{\text{TV}}\perp \overline{\text{PV}}\text{ and }\overline{\text{TV}}\perp \overline{\text{VS}}$ ……. if a line is perpendicular to plane than all line passing through its foot will also perpendicular to it

  $\angle \text{TVP}=\angle \text{TVS}=90°$ ……. perpendicular lines from right angles

  $\overline{\text{TV}}\cong \overline{\text{VS}}$ …… given

  $\overline{\text{PT}}\cong \overline{\text{PS}}$ …… reflexive property

  $\therefore \Delta \text{TVP}\cong \Delta \text{PVS}$

  $\overline{\text{TS}}\cong \overline{\text{PS}}$ ……. congruent parts of congruent triangles are congruent

  $\therefore \overline{\text{TS}}=\overline{\text{PS}}$ ,….. congruent parts are always equal

  $\therefore$ PV bisects the angle TPS

Hence proved