Chapter 6.2, Problem 14PSC

To determine

To prove: The given line segment EF is perpendicular to a given plane ‘m’

Explanation of Solution

Given information:

The given line segment $\left(\overline{\text{EF}}\right)$ is perpendicular to another given line segment $\left(\overline{\text{CF}}\right)$

The given line segment $\left(\overline{\text{CE}}\right)$ is congruent to another given line segment $\left(\overline{\text{DE}}\right)$

The given angle $\left(\angle \text{FCD}\right)$ is congruent to another given angle $\left(\angle \text{FDC}\right)$

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

Proof:

  $\overline{\text{EF}}\perp \overline{\text{CF}}$ .......... given

  $\overline{\text{EC}}\perp \overline{\text{CF}}$ .......... reflexive property

  $\angle \text{FCD}\cong \angle \text{FDC}$ .......... given

  $\therefore \angle \text{FCD=}\angle \text{FDC}$ .......... by definition

  $\overline{\text{ED}}=\overline{\text{ED}}$ .......... reflexive property

  $\therefore \Delta \text{EFC}\cong \Delta \text{EFD}$ .......... SAS criteria

  $\Delta \text{FCD}$ will lie on plane m and perpendicular to $\Delta \text{ECF}$ .......... as per definition

  $\therefore \overline{\text{EF}}\perp \text{m}$ .......... Congruent Parts of Congruent Triangles are Congruent

Hence proved