Chapter 6.2, Problem 17PSC

To determine

To prove: If a line is perpendicular to the plane of a circle and passing through center of the circle then any point on the line which is equidistant from any two points of the circle

Explanation of Solution

Given information:

A line given which is

perpendicular to the plane of a given circle and

passes through the circle’s center

  

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:

  $\overleftrightarrow{\text{OP}}\perp \text{m}$ ........... assume

  $\overline{\text{OA}}=\overline{\text{OB}}$ ........... radii of the circle

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

  $\angle \text{OAP}=\angle \text{OAB}=90°$ ........... Perpendicular lines from right angles

  $\overline{\text{OA}}\cong \overline{\text{OB}}$ ........... by definition

  $\Delta \text{OAP}\cong \Delta \text{OBP}$ ........... SAS criteria

  $\overline{\text{PA}}\cong \overline{\text{PB}}$ ........... Congruent Parts of Congruent Triangles are Congruent Reflexive property

  $\overline{\text{PA}}=\overline{\text{PB}}$ i.e., Points A and B are equidistance from external point P

Hence proved