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Math
GeometryElementary Geometry for College Students6th Edition

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Chapter 6.3, Problem 31E
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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

Solutions

Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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Does it follow from Exercise 30 that Δ A D E is also congruent to Δ C B E ? What can you conclude regarding A E ¯ and C E ¯ in the drawing? What can you conclude regarding D E ¯     a n d     B E ¯ ?

Chapter 6.3, Problem 31E, Does it follow from Exercise 30 that ADE is also congruent to CBE? What can you conclude regarding

To determine

To find:

To explain the ΔADE is also congruent to ΔCBE, also conclude regarding AE¯ and CE¯ in the drawing and regarding DE¯ and BE¯.

Explanation

Given that, AB¯≅CD¯ in ⊙P.

To prove ΔABD≅ΔCDB

The diagrammatic representation is given below,

From the above figure shows that ∠DAB,∠ABD,∠BCD,∠CDB have the vertex on the circle.

We know that inscribed angle is half of the arc measure. That is,

m∠DAB=mDB^2

m∠ABD=mAD^2

m∠BCD=mDB^2

m∠DAB=mBC^2

Given that AB¯≅CD¯ in ⊙P therefore, mADB^=mDBC^

Then add the arc to get the following,

mAD^+mDB^=mDB^+mBC^

mAD^=mBC^

By substituting the above

m∠DAB=mDB^2=m∠BCD

m∠DAB=m∠BCD

m∠ABD=mAD^2=mBC^2=m∠CDB

m∠ABD=m∠CDB

By using SAS postulate (two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent

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Sect-6.1 P-1ESect-6.1 P-2ESect-6.1 P-3ESect-6.1 P-4ESect-6.1 P-5ESect-6.1 P-6ESect-6.1 P-7ESect-6.1 P-8ESect-6.1 P-9ESect-6.1 P-10E
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