Explanation Given: A triangle with two equal angles. Calculation: Let A B C be a triangle with an angle bisector at A that meets BC at point D as shown in the figure below. Using two-column proof. Statements Reasons ∠ B = ∠ C Given ∠ B A D = ∠ C A D AD is the angle bisector of A A D = A D Reflexive property Δ A B D ≅ Δ A C D By AAS postulate A B = A C Using congruent parts of congruent triangles Therefore, A B = A C has been proved.

McDougal Littell Jurgensen Geometry: Student Edition Geometry

5th Edition

ISBN: 9780395977279

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

Publisher: Houghton Mifflin Company College Division

See similar textbooks

Concept explainers

Permutations and Combinations

If there are 5 dishes, they can be relished in any order at a time. In permutation, it should be in a particular order. In combination, the order does not matter. Take 3 letters a, b, and c. The possible ways of pairing any two letters are ab, bc, ac, ba…

Counting Theory

The fundamental counting principle is a rule that is used to count the total number of possible outcomes in a given situation.

Videos

Question

Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

Chapter 4.4, Problem 12WE

To determine

To Prove: If two angles of a triangle are congruent, then the sides opposite to these angles are congruent.

Expert Solution & Answer

Explanation of Solution

Given:

A triangle with two equal angles.

Calculation:

Let ABC be a triangle with an angle bisector at A that meets BC at point D as shown in the figure below.

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 4.4, Problem 12WE

Using two-column proof.

Statements	Reasons
∠B=∠C	Given
∠BAD=∠CAD	AD is the angle bisector of A
AD=AD	Reflexive property
ΔABD≅ΔACD	By AAS postulate
AB=AC	Using congruent parts of congruent triangles

Therefore, AB=AC has been proved.

Chapter 4.4, Problem 11WE

Chapter 4.4, Problem 13WE

Chapter 4 Solutions

McDougal Littell Jurgensen Geometry: Student Edition Geometry

Ch. 4.1 - Prob. 1CE Ch. 4.1 - Prob. 2CE Ch. 4.1 - Prob. 3CE Ch. 4.1 - Prob. 4CE Ch. 4.1 - Prob. 5CE Ch. 4.1 - Prob. 6CE Ch. 4.1 - Prob. 7CE Ch. 4.1 - Prob. 8CE Ch. 4.1 - Prob. 9CE Ch. 4.1 - Prob. 10CE

Ch. 4.1 - Prob. 11CE Ch. 4.1 - Prob. 12CE Ch. 4.1 - Prob. 13CE Ch. 4.1 - Prob. 14CE Ch. 4.1 - Prob. 15CE Ch. 4.1 - Prob. 16CE Ch. 4.1 - Prob. 17CE Ch. 4.1 - Prob. 18CE Ch. 4.1 - Prob. 19CE Ch. 4.1 - Prob. 20CE Ch. 4.1 - Prob. 1WE Ch. 4.1 - Prob. 2WE Ch. 4.1 - Prob. 3WE Ch. 4.1 - Prob. 4WE Ch. 4.1 - Prob. 5WE Ch. 4.1 - Prob. 6WE Ch. 4.1 - Prob. 7WE Ch. 4.1 - Prob. 8WE Ch. 4.1 - Prob. 9WE Ch. 4.1 - Prob. 10WE Ch. 4.1 - Prob. 11WE Ch. 4.1 - Prob. 12WE Ch. 4.1 - Prob. 13WE Ch. 4.1 - Prob. 14WE Ch. 4.1 - Prob. 15WE Ch. 4.1 - Prob. 16WE Ch. 4.1 - Prob. 17WE Ch. 4.1 - Prob. 18WE Ch. 4.1 - Prob. 19WE Ch. 4.1 - Prob. 20WE Ch. 4.1 - Prob. 21WE Ch. 4.1 - Prob. 22WE Ch. 4.1 - Prob. 23WE Ch. 4.1 - Prob. 24WE Ch. 4.1 - Prob. 1MRE Ch. 4.1 - Prob. 2MRE Ch. 4.2 - Prob. 1CE Ch. 4.2 - Prob. 2CE Ch. 4.2 - Prob. 3CE Ch. 4.2 - Prob. 4CE Ch. 4.2 - Prob. 5CE Ch. 4.2 - Prob. 6CE Ch. 4.2 - Prob. 7CE Ch. 4.2 - Prob. 8CE Ch. 4.2 - Prob. 9CE Ch. 4.2 - Prob. 10CE Ch. 4.2 - Prob. 11CE Ch. 4.2 - Prob. 1WE Ch. 4.2 - Prob. 2WE Ch. 4.2 - Prob. 3WE Ch. 4.2 - Prob. 4WE Ch. 4.2 - Prob. 5WE Ch. 4.2 - Prob. 6WE Ch. 4.2 - Prob. 7WE Ch. 4.2 - Prob. 8WE Ch. 4.2 - Prob. 9WE Ch. 4.2 - Prob. 10WE Ch. 4.2 - Prob. 11WE Ch. 4.2 - Prob. 12WE Ch. 4.2 - Prob. 13WE Ch. 4.2 - Prob. 14WE Ch. 4.2 - Prob. 15WE Ch. 4.2 - Prob. 16WE Ch. 4.2 - Prob. 17WE Ch. 4.2 - Prob. 18WE Ch. 4.2 - Prob. 19WE Ch. 4.2 - Prob. 20WE Ch. 4.2 - Prob. 21WE Ch. 4.2 - Prob. 22WE Ch. 4.2 - Prob. 23WE Ch. 4.2 - Prob. 24WE Ch. 4.2 - Prob. 25WE Ch. 4.2 - Prob. 26WE Ch. 4.2 - Prob. 27WE Ch. 4.3 - Prob. 1CE Ch. 4.3 - Prob. 2CE Ch. 4.3 - Prob. 3CE Ch. 4.3 - Prob. 4CE Ch. 4.3 - Prob. 5CE Ch. 4.3 - Prob. 6CE Ch. 4.3 - Prob. 1WE Ch. 4.3 - Prob. 2WE Ch. 4.3 - Prob. 3WE Ch. 4.3 - Prob. 4WE Ch. 4.3 - Prob. 5WE Ch. 4.3 - Prob. 6WE Ch. 4.3 - Prob. 7WE Ch. 4.3 - Prob. 8WE Ch. 4.3 - Prob. 9WE Ch. 4.3 - Prob. 10WE Ch. 4.3 - Prob. 11WE Ch. 4.3 - Prob. 12WE Ch. 4.3 - Prob. 13WE Ch. 4.3 - Prob. 14WE Ch. 4.3 - Prob. 15WE Ch. 4.3 - Prob. 16WE Ch. 4.3 - Prob. 1ST1 Ch. 4.3 - Prob. 2ST1 Ch. 4.3 - Prob. 3ST1 Ch. 4.3 - Prob. 4ST1 Ch. 4.3 - Prob. 5ST1 Ch. 4.3 - Prob. 6ST1 Ch. 4.3 - Prob. 7ST1 Ch. 4.3 - Prob. 8ST1 Ch. 4.4 - Prob. 1CE Ch. 4.4 - Prob. 2CE Ch. 4.4 - Prob. 3CE Ch. 4.4 - Prob. 4CE Ch. 4.4 - Prob. 5CE Ch. 4.4 - Prob. 6CE Ch. 4.4 - Prob. 7CE Ch. 4.4 - Prob. 8CE Ch. 4.4 - Prob. 9CE Ch. 4.4 - Prob. 10CE Ch. 4.4 - Prob. 11CE Ch. 4.4 - Prob. 12CE Ch. 4.4 - Prob. 1WE Ch. 4.4 - Prob. 2WE Ch. 4.4 - Prob. 3WE Ch. 4.4 - Prob. 4WE Ch. 4.4 - Prob. 5WE Ch. 4.4 - Prob. 6WE Ch. 4.4 - Prob. 7WE Ch. 4.4 - Prob. 8WE Ch. 4.4 - Prob. 9WE Ch. 4.4 - Prob. 10WE Ch. 4.4 - Prob. 11WE Ch. 4.4 - Prob. 12WE Ch. 4.4 - Prob. 13WE Ch. 4.4 - Prob. 14WE Ch. 4.4 - Prob. 15WE Ch. 4.4 - Prob. 16WE Ch. 4.4 - Prob. 17WE Ch. 4.4 - Prob. 18WE Ch. 4.4 - Prob. 19WE Ch. 4.4 - Prob. 20WE Ch. 4.4 - Prob. 21WE Ch. 4.4 - Prob. 22WE Ch. 4.4 - Prob. 23WE Ch. 4.4 - Prob. 24WE Ch. 4.4 - Prob. 25WE Ch. 4.4 - Prob. 26WE Ch. 4.4 - Prob. 27WE Ch. 4.4 - Prob. 28WE Ch. 4.4 - Prob. 29WE Ch. 4.4 - Prob. 30WE Ch. 4.4 - Prob. 31WE Ch. 4.4 - Prob. 32WE Ch. 4.4 - Prob. 33WE Ch. 4.4 - Prob. 34WE Ch. 4.5 - Prob. 1CE Ch. 4.5 - Prob. 2CE Ch. 4.5 - Prob. 3CE Ch. 4.5 - Prob. 4CE Ch. 4.5 - Prob. 5CE Ch. 4.5 - Prob. 6CE Ch. 4.5 - Prob. 7CE Ch. 4.5 - Prob. 8CE Ch. 4.5 - Prob. 9CE Ch. 4.5 - Prob. 10CE Ch. 4.5 - Prob. 11CE Ch. 4.5 - Prob. 12CE Ch. 4.5 - Prob. 13CE Ch. 4.5 - Prob. 14CE Ch. 4.5 - Prob. 1WE Ch. 4.5 - Prob. 2WE Ch. 4.5 - Prob. 3WE Ch. 4.5 - Prob. 4WE Ch. 4.5 - Prob. 5WE Ch. 4.5 - Prob. 6WE Ch. 4.5 - Prob. 7WE Ch. 4.5 - Prob. 8WE Ch. 4.5 - Prob. 9WE Ch. 4.5 - Prob. 10WE Ch. 4.5 - Prob. 11WE Ch. 4.5 - Prob. 12WE Ch. 4.5 - Prob. 13WE Ch. 4.5 - Prob. 14WE Ch. 4.5 - Prob. 15WE Ch. 4.5 - Prob. 16WE Ch. 4.5 - Prob. 17WE Ch. 4.5 - Prob. 18WE Ch. 4.5 - Prob. 19WE Ch. 4.5 - Prob. 20WE Ch. 4.5 - Prob. 21WE Ch. 4.5 - Prob. 22WE Ch. 4.5 - Prob. 1ST2 Ch. 4.5 - Prob. 2ST2 Ch. 4.5 - Prob. 3ST2 Ch. 4.5 - Prob. 4ST2 Ch. 4.5 - Prob. 5ST2 Ch. 4.6 - Prob. 1CE Ch. 4.6 - Prob. 2CE Ch. 4.6 - Prob. 3CE Ch. 4.6 - Prob. 4CE Ch. 4.6 - Prob. 1WE Ch. 4.6 - Prob. 2WE Ch. 4.6 - Prob. 3WE Ch. 4.6 - Prob. 4WE Ch. 4.6 - Prob. 5WE Ch. 4.6 - Prob. 6WE Ch. 4.6 - Prob. 7WE Ch. 4.6 - Prob. 8WE Ch. 4.6 - Prob. 9WE Ch. 4.6 - Prob. 10WE Ch. 4.6 - Prob. 11WE Ch. 4.6 - Prob. 12WE Ch. 4.6 - Prob. 13WE Ch. 4.6 - Prob. 14WE Ch. 4.6 - Prob. 15WE Ch. 4.6 - Prob. 16WE Ch. 4.6 - Prob. 17WE Ch. 4.6 - Prob. 1MRE Ch. 4.6 - Prob. 2MRE Ch. 4.6 - Prob. 3MRE Ch. 4.6 - Prob. 4MRE Ch. 4.6 - Prob. 5MRE Ch. 4.6 - Prob. 6MRE Ch. 4.6 - Prob. 7MRE Ch. 4.6 - Prob. 8MRE Ch. 4.6 - Prob. 9MRE Ch. 4.6 - Prob. 10MRE Ch. 4.6 - Prob. 11MRE Ch. 4.7 - Prob. 1CE Ch. 4.7 - Prob. 2CE Ch. 4.7 - Prob. 3CE Ch. 4.7 - Prob. 4CE Ch. 4.7 - Prob. 5CE Ch. 4.7 - Prob. 6CE Ch. 4.7 - Prob. 7CE Ch. 4.7 - Prob. 8CE Ch. 4.7 - Prob. 9CE Ch. 4.7 - Prob. 10CE Ch. 4.7 - Prob. 11CE Ch. 4.7 - Prob. 1WE Ch. 4.7 - Prob. 2WE Ch. 4.7 - Prob. 3WE Ch. 4.7 - Prob. 4WE Ch. 4.7 - Prob. 5WE Ch. 4.7 - Prob. 6WE Ch. 4.7 - Prob. 7WE Ch. 4.7 - Prob. 8WE Ch. 4.7 - Prob. 9WE Ch. 4.7 - Prob. 10WE Ch. 4.7 - Prob. 11WE Ch. 4.7 - Prob. 12WE Ch. 4.7 - Prob. 13WE Ch. 4.7 - Prob. 14WE Ch. 4.7 - Prob. 15WE Ch. 4.7 - Prob. 16WE Ch. 4.7 - Prob. 17WE Ch. 4.7 - Prob. 18WE Ch. 4.7 - Prob. 19WE Ch. 4.7 - Prob. 20WE Ch. 4.7 - Prob. 21WE Ch. 4.7 - Prob. 22WE Ch. 4.7 - Prob. 23WE Ch. 4.7 - Prob. 24WE Ch. 4.7 - Prob. 25WE Ch. 4.7 - Prob. 26WE Ch. 4.7 - Prob. 27WE Ch. 4.7 - Prob. 28WE Ch. 4.7 - Prob. 29WE Ch. 4.7 - Prob. 1ST3 Ch. 4.7 - Prob. 2ST3 Ch. 4.7 - Prob. 3ST3 Ch. 4.7 - Prob. 4ST3 Ch. 4.7 - Prob. 5ST3 Ch. 4.7 - Prob. 6ST3 Ch. 4 - Prob. 1CR Ch. 4 - Prob. 2CR Ch. 4 - Prob. 3CR Ch. 4 - Prob. 4CR Ch. 4 - Prob. 5CR Ch. 4 - Prob. 6CR Ch. 4 - Prob. 7CR Ch. 4 - Prob. 8CR Ch. 4 - Prob. 9CR Ch. 4 - Prob. 10CR Ch. 4 - Prob. 11CR Ch. 4 - Prob. 12CR Ch. 4 - Prob. 13CR Ch. 4 - Prob. 14CR Ch. 4 - Prob. 15CR Ch. 4 - Prob. 16CR Ch. 4 - Prob. 17CR Ch. 4 - Prob. 18CR Ch. 4 - Prob. 19CR Ch. 4 - Prob. 20CR Ch. 4 - Prob. 1CT Ch. 4 - Prob. 2CT Ch. 4 - Prob. 3CT Ch. 4 - Prob. 4CT Ch. 4 - Prob. 5CT Ch. 4 - Prob. 6CT Ch. 4 - Prob. 7CT Ch. 4 - Prob. 8CT Ch. 4 - Prob. 9CT Ch. 4 - Prob. 10CT Ch. 4 - Prob. 11CT Ch. 4 - Prob. 12CT Ch. 4 - Prob. 13CT Ch. 4 - Prob. 14CT Ch. 4 - Prob. 15CT Ch. 4 - Prob. 16CT Ch. 4 - Prob. 17CT Ch. 4 - Prob. 18CT Ch. 4 - Prob. 19CT Ch. 4 - Prob. 20CT Ch. 4 - Prob. 1AR Ch. 4 - Prob. 2AR Ch. 4 - Prob. 3AR Ch. 4 - Prob. 4AR Ch. 4 - Prob. 5AR Ch. 4 - Prob. 6AR Ch. 4 - Prob. 7AR Ch. 4 - Prob. 8AR Ch. 4 - Prob. 9AR Ch. 4 - Prob. 10AR Ch. 4 - Prob. 11AR Ch. 4 - Prob. 12AR Ch. 4 - Prob. 13AR Ch. 4 - Prob. 14AR Ch. 4 - Prob. 15AR Ch. 4 - Prob. 16AR Ch. 4 - Prob. 17AR Ch. 4 - Prob. 18AR Ch. 4 - Prob. 19AR Ch. 4 - Prob. 20AR Ch. 4 - Prob. 21AR Ch. 4 - Prob. 22AR Ch. 4 - Prob. 23AR Ch. 4 - Prob. 24AR Ch. 4 - Prob. 25AR Ch. 4 - Prob. 26AR Ch. 4 - Prob. 27AR Ch. 4 - Prob. 28AR Ch. 4 - Prob. 29AR Ch. 4 - Prob. 30AR Ch. 4 - Prob. 31AR Ch. 4 - Prob. 32AR Ch. 4 - Prob. 33AR Ch. 4 - Prob. 1CUR Ch. 4 - Prob. 2CUR Ch. 4 - Prob. 3CUR Ch. 4 - Prob. 4CUR Ch. 4 - Prob. 5CUR Ch. 4 - Prob. 6CUR Ch. 4 - Prob. 7CUR Ch. 4 - Prob. 8CUR Ch. 4 - Prob. 9CUR Ch. 4 - Prob. 10CUR Ch. 4 - Prob. 11CUR Ch. 4 - Prob. 12CUR Ch. 4 - Prob. 13CUR Ch. 4 - Prob. 14CUR Ch. 4 - Prob. 15CUR Ch. 4 - Prob. 16CUR Ch. 4 - Prob. 17CUR Ch. 4 - Prob. 18CUR

Additional Math Textbook Solutions

Find more solutions based on key concepts

How many outcome sequences are possible ten a die is rolled four times, where we say, for instance, that the ou...

A First Course in Probability (10th Edition)

In Problems 37-48, determine algebraically whether each function is even, odd, or neither. 43. f( x )=x+| x |

Precalculus

CHECK POINT I Let p and q represent the following statements: p : 3 + 5 = 8 q : 2 × 7 = 20. Determine the truth...

Thinking Mathematically (6th Edition)

Evaluate the integrals in Exercises 17–66. 21.

University Calculus: Early Transcendentals (4th Edition)

The table by using the given graph of h.

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

Identifying Type I and Type II Errors In Exercises 31–36, describe type I and type II errors for a hypothesis t...

Elementary Statistics: Picturing the World (7th Edition)

Knowledge Booster

$Geometry$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, geometry and related others by exploring similar questions and additional content below.

Recommended textbooks for you

Elementary Geometry For College Students, 7e
Geometry
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Cengage,
Elementary Geometry for College Students
Geometry
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Cengage Learning

Elementary Geometry For College Students, 7e

Geometry

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Cengage,

Elementary Geometry for College Students

Geometry

ISBN:9781285195698

Author:Daniel C. Alexander, Geralyn M. Koeberlein

Publisher:Cengage Learning

SEE MORE TEXTBOOKS

Find number of persons in a part with 66 handshakes Combinations; Author: Anil Kumar;https://www.youtube.com/watch?v=33TgLi-wp3E;License: Standard YouTube License, CC-BY

Discrete Math 6.3.1 Permutations and Combinations; Author: Kimberly Brehm;https://www.youtube.com/watch?v=J1m9sB5XZQc;License: Standard YouTube License, CC-BY

How to use permutations and combinations; Author: Mario's Math Tutoring;https://www.youtube.com/watch?v=NEGxh_D7yKU;License: Standard YouTube License, CC-BY

Permutations and Combinations | Counting | Don't Memorise; Author: Don't Memorise;https://www.youtube.com/watch?v=0NAASclUm4k;License: Standard Youtube License

Permutations and Combinations Tutorial; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=XJnIdRXUi7A;License: Standard YouTube License, CC-BY

To Prove : If two angles of a triangle are congruent, then the sides opposite to these angles are congruent.