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Math
GeometryElementary Geometry for College Students6th Edition

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Chapter 6.3, Problem 28E
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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
BuyFindarrow_forward

Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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In Exercises 27 to 30, provide a paragraph proof.

Given: ⊙ O   with O M ¯ ⊥ A B ¯     a n d     O N ¯ ⊥ B C ¯

O M ¯ ≅ O N ¯ Secant A C   ¯   t o     ⊙ O

Secant A E   ¯   t o     ⊙ Q

Prove: Δ A B C is isosceles.

Chapter 6.3, Problem 28E, In Exercises 27 to 30, provide a paragraph proof. Given:O with OMABandONBC OMON Secant ACtoO Secant

To determine

To find:

To prove ΔABC is isosceles.

Explanation

Given that, ⊙O with OM¯⊥AB¯ and ON¯⊥BC¯ and OM¯≅ON¯.

To prove ΔABC is isosceles that is an isosceles triangle is a triangle that has two sides of equal length.

The diagrammatic representation is given below,

First compare the ΔONB,ΔOMB

But OB=OB. Since its common for both triangles and identity.

Given that OM¯⊥AB¯ and ON¯⊥BC¯

Therefore, ∠OMB=∠ONB=90∘

(SAS) Definition: Triangles are congruent if any pair of corresponding sides and their included angles are equal in both triangles.

So by using side angle side property ΔOMB≅ΔONB.

Also using CPCTC (corresponding parts of congruent triangles are congruent) to get the following,

MB¯=NB¯...(1)

Theorem 6.3.1:

If a line is drawn through the center of a circle perpendicular to a chord, then it bisects the chord and its arc.

That is for circle ⊙O given OM¯⊥AB¯, therefore AM¯=MB¯...(2)

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