Chapter 6.2, Problem 15PSC

To determine

To prove: The two given line segments AD and BC are congruent

Explanation of Solution

Given information:

Two given lines $\left(\overleftrightarrow{\text{AD}}\&\overleftrightarrow{\text{BC}}\right)$ intersect at one point (E)

One given line $\left(\overleftrightarrow{\text{AC}}\right)$ is perpendicular to the given planes (m & n)

Another given line $\left(\overleftrightarrow{\text{BD}}\right)$ is perpendicular to the given planes (m & n)

Formula used:

Congruency means same shape and same size

Congruent Parts of Congruent Triangles are Congruent

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

Proof:

  $\overleftrightarrow{\text{AC}}\perp \text{m \& n}$ and $\overleftrightarrow{\text{BD}}\perp \text{m \& n}$ .......... given

  $\angle \text{ACD=}\angle \text{BDC=90°}$ ........... perpendicular lines from right angles

  $\overline{\text{AC}}\cong \overline{\text{BD}}$ ........... $\overleftrightarrow{\text{AC}}\perp \text{m \& n}$ and $\overleftrightarrow{\text{BD}}\perp \text{m \& n}$ , given

  $\overline{\text{CD}}\cong \overline{\text{AB}}$ ............ reflexive property

  $\therefore \Delta \text{ACD}\cong \Delta \text{BDC}$ .......... SAS criteria

  $\overline{\text{AD}}\cong \overline{\text{BC}}$ .......... Congruent Parts of Congruent Triangles are Congruent

Hence proved