Chapter 6.2, Problem 13PSB

To determine

To prove: The oblique segments are congruent, if the foot of the perpendicular is equidistant from the feet of the oblique lines

Explanation of Solution

Given information:

Any point (P) on a plane (k)

From any point (P) line perpendicular (PA) to a plane (k), two lines (PG & PH) are drawn

Lines (PG & PH) drawn are oblique to the plane (k)

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

Right angles are congruent, by definition of congruency

Proof:

  $\overleftrightarrow{\text{PA}}\perp \text{k}$ …….. given

  $\overline{\text{AH}}\perp \text{k}$ & $\overline{\text{AG}}\perp \text{k}$ …… If a line is perpendicular to plane than all line passing through its foot will also perpendicular to it

  $\angle \text{PAG}=\angle \text{PAH}=90°$ …… $\overleftrightarrow{\text{PA}}\perp \text{k}$ is given

  $\angle \text{PAG}\cong \angle \text{PAH}$ ….. Right angles are congruent, by definition of congruency

  $\therefore \Delta \text{PAG}\cong \Delta \text{PAH}$ …… ASA criteria

  $\overline{\text{PA}}\cong \overline{\text{PB}}$ …….. congruent Parts of congruent triangles are congruent

Hence proved