Q25. For Exercises 23-27 write proofs... [FREE SOLUTION]

91影视

Geometry

Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

$Math Studyset 91影视 Explanations$ Math

Student Edition 路 227 pages

ISBN: 9780395977279

Chapter 4: Q25. (page 158)

For Exercises 23-27 write proofs in paragraph form. (Hint: You can use theorems from this section to write fairly short proofs for Exercises 23 and 24.)
Given: Plane $M$ is the perpendicular bisecting plane of $\overset{炉}{A B}$ , (That is, $\overset{炉}{A B} 鈯�$ plane $M$ and $O$ is the midpoint of $\overset{炉}{A B}$ )
Prove: $a . \overset{炉}{A D} 鈮� \overset{炉}{B D} b . \overset{炉}{A C} 鈮� \overset{炉}{B C} c . 鈭� C A D 鈮� 鈭� C B D$

Short Answer

Expert verified

Point $D$ lies on the plane $M$ which is perpendicular bisector for line segment $\overset{炉}{A B}$ and $O$ is its midpoint. So $\overset{炉}{O D}$ is perpendicular bisector of $\overset{炉}{A B}$ which implies point $D$ is equidistance from endpoints $A$ and $B$ . Thus $\overset{炉}{A D} 鈮� \overset{炉}{B D}$
Point $C$ lies on the plane $M$ which is perpendicular bisector for line segment $\overset{炉}{A B}$ and $O$ is its midpoint. So $\overset{炉}{O C}$ is perpendicular bisector of $\overset{炉}{A B}$ which implies point $C$ is equidistance from endpoints $A$ and $B$ . Thus $\overset{炉}{A C} 鈮� \overset{炉}{B C}$
In width="48" height="19" role="math">螖CADand $螖 C B D$ , using above two parts $\overset{炉}{A D} 鈮� \overset{炉}{B D}$ and $\overset{炉}{A C} 鈮� \overset{炉}{B C}$ . Also, $\overset{炉}{C D}$ is common in both triangles, so by reflexive property it is congruent to itself. Thus, by SSS (Side-Side-Side) Postulate $螖 C A D 鈮� 螖 C B D$ Since, corresponding parts of congruent triangles are congruent, so $鈭� C A D 鈮� 鈭� C B D$

Step by step solution

Part a. Step 1. Observation from image.

Point $D$ lies on the plane $M$ which is perpendicular bisector for line segment $\overset{炉}{A B}$ and $O$ is its midpoint. So $\overset{炉}{O D}$ is perpendicular bisector of $\overset{炉}{A B}$

Part a. Step 2. Show that AD¯≅BD¯.

Since, $\overset{炉}{O D}$ is perpendicular bisector of $\overset{炉}{A B}$ , so point $D$ is equidistance from endpoints $A$ and $B$ . Thus $\overset{炉}{A D} 鈮� \overset{炉}{B D}$

Part b. Step 1. Observation from image.

Point $C$ lies on the plane $M$ which is perpendicular bisector for line segment $\overset{炉}{A B}$ and $O$ is its midpoint. So $\overset{炉}{O C}$ is perpendicular bisector of $\overset{炉}{A B}$

Part b. Step 2. Show that AD¯≅BD¯.

Since, $\overset{炉}{O C}$ is perpendicular bisector of $\overset{炉}{A B}$ , so point $C$ is equidistance from endpoints $A$ and $B$ . Thus $\overset{炉}{A C} 鈮� \overset{炉}{B C}$