Chapter 6.2, Problem 9PSB

To determine

To prove: The given point P is equidistant from other two given points, i.e., B and C

Explanation of Solution

Given information:

Given line segment is perpendicular to plane, i.e., $\overline{\text{PA}}\perp s$ and

Given point (P) is equidistant from B and C

Formula used:

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 be perpendicular to it

Perpendicular lines from right angles are always congruent if any one of them lie in a given plane

Proof:

  $\overline{\text{PA}}\perp s$ ........... given

Points P is equidistant from B and C B........... given

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

  $\overline{\text{PA}}\cong \overline{\text{PB}}$ ........... Perpendicular lines from right angles

  $\overline{\text{PA}}\cong \overline{\text{PC}}$ ........... reflexive

  $\therefore$$\Delta \text{PAB}\cong \Delta \text{PBC}$

A is equidistant from B and C.......... congruent Parts of congruent triangles are congruent

Hence proved