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 Post subject: GMAT Geometry
PostPosted: Mon May 13, 2013 7:52 am 
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Joined: Tue Apr 13, 2010 8:48 am
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In triangle ABC, DB and DC are angle bisectors and the angle BAC = 60°. If angle DCB is 40°, what is the degree measure of angle BDC? (Note: Figure not drawn to scale.)

A. 130°
B. 120°
C. 100°
D. 80°
E. 75°

(B) First of all remember that the sum of all angles in a triangle is equal to 180°. We are told that angle BAC = 60°. That leaves 120° to be split between angles ABC and ACB. We know that DC and DB are angle bisectors (an angle bisector can be described as a line segment that divides an angle into two equal angles), so the sum of angles DBC and DCB is 120°/2 = 60°. BDC is the third angle in triangle BDC. Therefore it equals 180° – 60° = 120°. The correct answer is B.
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I do not understand why the answer is 120° and not 100°. If as you say, DB and DC are angle bisectors and DCB is 40° degrees (as stated in the question) CBD should also be 40° (which lets CBA and ACB as 60 degrees angles). If all the angles must be 180 degrees 180° – 80° = 100°.
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 Post subject: Re: GMAT Geometry
PostPosted: Mon May 13, 2013 7:59 am 
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Joined: Fri Apr 09, 2010 2:11 pm
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CBD should also be 40° (which lets CBA and ACB as 60 degrees angles).
Angles CBA and ACB are NOT 60° angles.
An angle bisector divides that angle into two equal angles. So if angle DCB is 40° then angle ACB is 80°. You can calculate angle CBA, it equals 180° – 60° – 80° = 40°.
So angle CBD is NOT 40°. It equals 40°/2 = 20°.


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