Geometry For Enjoyment And Challenge
Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
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Chapter 6.1, Problem 16PSC
To determine

To show that side ST is perpendicular to side BD.

Expert Solution & Answer
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Explanation of Solution

Given information:

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

  ST intersects m at B.

D is any point on AC¯. .

  ST AB and BC  SB¯    TB¯

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,

  ST AB and BC  SB¯    TB¯

  Geometry For Enjoyment And Challenge, Chapter 6.1, Problem 16PSC

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

  AB¯ is perpendicular to bisector of ST¯CB¯ is perpendicular to bisector of ST¯

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

  AS¯   AT¯

  CS¯   CT¯

By reflexive property, we get

  AC¯   AC¯

By SSS congruence rule, we get

  ΔSAC    ΔTAC  

As corresponding parts of congruent triangles are congruent, we get

  SAD   TAD

By reflexive property, we get

  AD¯   AD¯

By SAS congruence rule, we get

  ΔSAD    ΔTAD  

As corresponding parts of congruent triangles are congruent, we get

  DS¯   DT¯

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

  DB¯ is perpendicular to bisector of ST¯

According to definition of perpendicular bisector

  ST¯BD

Chapter 6 Solutions

Geometry For Enjoyment And Challenge

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