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Measure that angle III (Posted on 2004-09-07) Difficulty: 4 of 5
Triangle ABC has side AB=AC, and angle BAC = 20 degrees.

D is a point on side AC with AD=BC. Find angle DBC.

Solve this without trigonometry.










See The Solution Submitted by Brian Smith    
Rating: 3.2000 (5 votes)

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No Subject | Comment 20 of 25 |

It's true--E must be to the upper left of the original triangle, otherwise BAE wouldn't be 100, as asserted.

Why are the angles at F 60 and 120? Notice that Triangle AEB is isoceles. (AB=AE).  So, the base angles must be 40, since we already know that the vertex at A is 100.  Angle AED is 20, so DEF must be 20.  EDF is 100; that leaves DFE = 60.  So AFB is 120.

I also noticed that AE and CB are parallel, but I don't know where that gets me.

Jim's assertion that FB = FD is what I'm having trouble with... I know it's true--I just don't see the proof. 


  Posted by Ken Haley on 2004-09-09 23:31:05
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