Chapter 6.1, Problem 16PSC

To determine

To show that side ST is perpendicular to side BD.

Explanation of Solution

Given information:

A, B, C and D lie in plane m .

  $\overleftrightarrow{ST}$ intersects m at B.

D is any point on $\overline{AC}.$ .

  $\begin{array}{l}\overleftrightarrow{ST}\perp \overleftrightarrow{AB}and\overleftrightarrow{BC} \\ \overline{SB} \cong \overline{TB}\end{array}$

Formula used:

The below properties are used:

If a segment divides a segment into two congruent segments, it is a bisector.

A point on a perpendicular bisector of a segment is equal to distance from ends of the segments.

Two triangles are congruent by SSS congruence rule.

Two angles are congruent by SAS congruence rule.

A point equal distance from endpoints of a segment is on the perpendicular bisector of the segment.

Proof:

It is given that,

  $\begin{array}{l}\overleftrightarrow{ST}\perp \overleftrightarrow{AB}and\overleftrightarrow{BC} \\ \overline{SB} \cong \overline{TB}\end{array}$

  

If a segment divides a segment into two congruent segments, it is a bisector.

  $\begin{array}{l}\overline{AB}isperpendiculartobisectorof\overline{ST}\hfill \\ \overline{CB}isperpendiculartobisectorof\overline{ST}\hfill \end{array}$

A point on a perpendicular bisector of a segment is equal to distance from ends of the segments.

  $\overline{AS} \cong \overline{AT}$

  $\overline{CS} \cong \overline{CT}$

By reflexive property, we get

  $\overline{AC} \cong \overline{AC}$

By SSS congruence rule, we get

  $\Delta SAC \cong \Delta TAC$

As corresponding parts of congruent triangles are congruent, we get

  $\angle SAD\cong \angle TAD$

By reflexive property, we get

  $\overline{AD} \cong \overline{AD}$

By SAS congruence rule, we get

  $\Delta SAD \cong \Delta TAD$

As corresponding parts of congruent triangles are congruent, we get

  $\overline{DS}\cong \overline{DT}$

A point equal distance from endpoints of a segment is on the perpendicular bisector of the segment.

  $\overline{DB}isperpendiculartobisectorof\overline{ST}$

According to definition of perpendicular bisector

  $\overline{ST}\perp \overrightarrow{BD}$