Chapter 6.2, Problem 5PSA

To determine

To prove: T is the midpoint of the AB and CD

Explanation of Solution

Given information:

Two given lines lie in a plane given, i.e., $\overleftrightarrow{\text{AB}}$ and $\overleftrightarrow{\text{CD}}$ lie in plane s

Given line is perpendicular to the given plane, i.e., $\overleftrightarrow{\text{PT}}\perp s$

Two given line segments are congruent, i.e., $$

  $\overline{\text{PC}}\cong \overline{\text{PD}}$

Two given line segments are congruent, i.e., $$

  $\overline{\text{PA}}\cong \overline{\text{PB}}$

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:

  $\overleftrightarrow{\text{PT}}\perp s$ …… given

  $\overline{\text{PT}}\perp \overline{\text{AB}}\text{ \& }\overline{\text{PT}}\perp \overline{\text{CD}}$ ……. if a line is perpendicular to plane than all line passing through its foot will also perpendicular to it

  $\angle \text{PTB}=\angle \text{PTA}=90°\&\angle \text{PTC}=\angle \text{PTD}=90°$ ….. Perpendicular lines from right angles

  $\overline{\text{PC}}\cong \overline{\text{PD}}\text{ \& }\overline{\text{PA}}\cong \overline{\text{PB}}$ ……. given

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

  $\therefore \Delta \text{PAB}\cong \Delta \text{PCD}$ …… SAS criteria

  $\overline{\text{TA}}\cong \overline{\text{TS}}\text{ \&}\overline{\text{TC}}\cong \overline{\text{TD}}$ ……. Congruent Parts of congruent Triangles are Congruent

  $\therefore$ T is the midpoint of $\overline{\text{AB}}\&\overline{\text{CD}}$

Hence proved