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

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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 7.4, Problem 27WE

To determine

To Prove: that AE⋅DG=AG⋅HE in the figure shown such that AH→⊥EH→ and AD→⊥DG→ .

Expert Solution & Answer

Explanation of Solution

Given:

Figure is shown as.

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 7.4, Problem 27WE

AH→⊥EH→ and AD→⊥DG→

Calculation:

In both triangles AHE and ADG

∠ADH=∠AHE ∴(Right angles)∠A=∠A ∴(Common)

Therefore, by AA postulate ΔAHE∼ΔADG

Ratio of sides of two similar triangle is equal.

AEAG=HEDGAH×DG=AG×DGAH⋅DG=AG⋅DG

So, it has been proved.

Chapter 7.4, Problem 26WE

Chapter 7.4, Problem 28WE

Chapter 7 Solutions

McDougal Littell Jurgensen Geometry: Student Edition Geometry

Ch. 7.1 - Prob. 1CE Ch. 7.1 - Prob. 2CE Ch. 7.1 - Prob. 3CE Ch. 7.1 - Prob. 4CE Ch. 7.1 - Prob. 5CE Ch. 7.1 - Prob. 6CE Ch. 7.1 - Prob. 7CE Ch. 7.1 - Prob. 8CE Ch. 7.1 - Prob. 9CE Ch. 7.1 - Prob. 10CE

Ch. 7.1 - Prob. 11CE Ch. 7.1 - Prob. 12CE Ch. 7.1 - Prob. 13CE Ch. 7.1 - Prob. 14CE Ch. 7.1 - Prob. 15CE Ch. 7.1 - Prob. 16CE Ch. 7.1 - Prob. 17CE Ch. 7.1 - Prob. 18CE Ch. 7.1 - Prob. 1WE Ch. 7.1 - Prob. 2WE Ch. 7.1 - Prob. 3WE Ch. 7.1 - Prob. 4WE Ch. 7.1 - Prob. 5WE Ch. 7.1 - Prob. 6WE Ch. 7.1 - Prob. 7WE Ch. 7.1 - Prob. 8WE Ch. 7.1 - Prob. 9WE Ch. 7.1 - Prob. 10WE Ch. 7.1 - Prob. 11WE Ch. 7.1 - Prob. 12WE Ch. 7.1 - Prob. 13WE Ch. 7.1 - Prob. 14WE Ch. 7.1 - Prob. 15WE Ch. 7.1 - Prob. 16WE Ch. 7.1 - Prob. 17WE Ch. 7.1 - Prob. 18WE Ch. 7.1 - Prob. 19WE Ch. 7.1 - Prob. 20WE Ch. 7.1 - Prob. 21WE Ch. 7.1 - Prob. 22WE Ch. 7.1 - Prob. 23WE Ch. 7.1 - Prob. 24WE Ch. 7.1 - Prob. 25WE Ch. 7.1 - Prob. 26WE Ch. 7.1 - Prob. 27WE Ch. 7.1 - Prob. 28WE Ch. 7.1 - Prob. 29WE Ch. 7.1 - Prob. 30WE Ch. 7.1 - Prob. 31WE Ch. 7.1 - Prob. 32WE Ch. 7.1 - Prob. 33WE Ch. 7.1 - Prob. 34WE Ch. 7.1 - Prob. 35WE Ch. 7.1 - Prob. 36WE Ch. 7.2 - Prob. 1CE Ch. 7.2 - Prob. 2CE Ch. 7.2 - Prob. 3CE Ch. 7.2 - Prob. 4CE Ch. 7.2 - 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To Prove: that A E ⋅ D G = A G ⋅ H E in the figure shown such that A H → ⊥ E H → and A D → ⊥ D G → .

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To Prove: that AE⋅DG=AG⋅HE in the figure shown such that AH→⊥EH→ and AD→⊥DG→ .

Explanation of Solution

Chapter 7 Solutions

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