Chapter 6.2, Problem 18PSC

To determine

To prove: The two triangles PAB and PDC are congruent

Explanation of Solution

Given information:

An isosceles trapezoid ABCD in a given plane t

Two-line segments are parallel to each other, i.e., $\overline{\text{BC}}\parallel \overline{\text{AD}}$

Line segment given is perpendicular to given plane, i.e., $\overline{\text{PF}}\perp \text{t}$

Line segment $\left(\overline{PF}\right)$ given bisects the other given line segment $\left(\overline{\text{AD}}\right)$

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:

Two-line segments are parallel to each other, i.e., $\overline{\text{BC}}\parallel \overline{\text{AD}}$

Line segment given is perpendicular to given plane, i.e., $\overline{\text{PF}}\perp \text{t}$

Line segment $\left(\overline{PF}\right)$ given bisects the other given line segment $\left(\overline{\text{AD}}\right)$

  $\overline{\text{BC}}\parallel \overline{\text{AD}}$ and $\overline{\text{PF}}\perp \text{t}$ ........... Given

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

  $\angle \text{PFA}=\angle \text{PFD}=90°$ ........... Perpendicular lines from right angles

  $\overline{\text{AP}}\cong \overline{\text{PC}}$ and $\overline{\text{PB}}\perp \overline{\text{PD}}$ ........... reflexive property

  $\overline{\text{AB}}\cong \overline{\text{CD}}$ ........... sides of an isosceles triangle

  $\Delta \text{PAB}\cong \Delta \text{PDC}$ ........... SSS criteria

Hence proved