geometry problem: find the lengths of DB, AE, and DE

-Fred151-

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This one is from Geometry by Harold R. Jacobs, page 638 numbers 58 through 60.

http://img146.imageshack.us/img146/8895 ... lemkl4.png

Black and blue were part of the image in the book; red is my assumtions so far. I'm not sure if F is useful or not.

I'm supposed to find the lengths of DB, AE, and DE.
 

pka

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I wish you had left off your additions.

What is the complete information set given with the problem?
 

stapel

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-Fred151-

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Look at the unedited image.
 

stapel

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-Fred151- said:
Look at the unedited image.
Since the "unedited" image includes hand-drawn bits, it isn't likely the image from the book. Also, the image does not include any instructions; were it the actual exercise, you would have had no idea what you needed to do. So you must have been provided with something else.

Please reply with the full and complete text of the exercise as it was provided to you. Thank you.

Eliz.
 

-Fred151-

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Exact words:

"In [triangle]ABC, [fay]AD bisects [angle]CAB and [line]DA[is parallel to][line]CA; AC = 8, CK = 6, and AB = 12."

I have no scanner. :cry:
 

galactus

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It would appear \(\displaystyle {\Delta}DBA\) is isosceles. Ergo, DB=12.
 

Denis

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Disagree galactus; DB = 9, AE = 4 4/5, DE = 4 4/5

Fred, add line DF parallel to AB, F on AC;
that gives you parallelogram AEDF, with all sides equal;
triangle CDF is similar to triangle ABC, so is triangle DBE;
from that, BC easily calculated to equal 15, then sides of AEDF to equal 4 4/5.

Draw your diagram more to scale: angle B = ~32, angle A = ~95;
then it'll be easier to "see" your way through the calculations...
 

galactus

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Denis said:
Disagree galactus;
No!, I'm right!..... (I am putting my fingers in my ears and humming real loud) :lol: :wink: :D
 
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