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Chapter 5.1, Problem 42E
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### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

#### Solutions

Chapter
Section
BuyFindarrow_forward

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

# In the figure, assume that a > 0. Prove that B D B A ≠ B E B C .

To determine

To prove:

BDBABEBC.

Explanation

It is given, BD = a, DA = 1, BE = a + 4 and EC = 3.

Find BA and BC using BD = a, DA = 1, BE = a + 4 and EC = 3.

BA = BD + DA

= a + 1

BC= BE + EC

= a + 4 + 3

= a + 7

Substitute BD = a and BA = a + 1 in BDBA.

BDBA=aa+1Â Â Â Â Â â€‰.....(1)

Substitute BE = a + 4 and BC = a + 7 in BEBC.

BEBC=a+4a+7Â Â â€‰

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