Chapter 6.2, Problem 11PSB

To determine

To prove: The triangle ACD is an isosceles triangle

Explanation of Solution

Given information:

The line segment (AB) is perpendicular to a given plane (m), i.e., $\overleftrightarrow{\text{AB}}\perp \text{m}$

An equilateral triangle $\left(\Delta \text{ABC}\right)$ lies in a given plane (m)

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

SAS property of congruence

Proof:

  $\overline{\text{AB}}\perp \text{m}$ …… given

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

  $\angle \text{ACD}=\angle \text{ABD}=90°$ …. Perpendicular lines from right angles

  $\overline{\text{BC}}\cong \overline{\text{BD}}$ ….. triangle DBC is an equilateral triangle

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

  $\therefore \Delta \text{ABC}\cong \Delta \text{ABD}$ ….. SAS property

  $\overline{\text{AD}}\cong \overline{\text{AC}}$ ….. Congruent Parts of congruent Triangles are Congruent

  $\overline{\text{AD}}=\overline{\text{AC}}$ …. Congruent sides are always equal

  $\therefore \Delta \text{ACD}$ is an isosceles triangle

Hence Proved