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Find angle DEB


Date: 5/27/96 at 19:52:6
From: Anonymous
Subject: Find angle DEB

Given an isosceles triangle ABC...
Angle BAC = 20 degrees and angle ABC = angle ACB = 80 degrees.
Draw a straight line from point C to point D on the other side of
the triangle. Angle BCD is 50 degrees. Similarly, a straight line
is drawn from point B to point E on the other side of the triangle. 
Angle CBE is 60 degrees. Points D and E are joined. The problem is
to find angle DEB. I have solved this using trigonometry. (Angle DEB
= 30 degrees.) However, I cannot derive a pure Euclidean solution. I 
have tried using similar triangles. I have tried inscribing the 
triangle in a circle, and I have tried constructing a right angled
triangle with BC as the base in trying to find some relationship that
will give me a solution, without success. I would appreciate any
help that you could give me. 


Date: 5/31/96 at 15:16:50
From: Doctor Ceeks
Subject: Re: Find angle DEB

This problem can be solved without reference to trigonometry as 
follows:

Take a circle and mark off points on the circumference every 20 
degrees to form the vertices of a regular 18-gon.  Let A be the center 
of the circle, and let B and C be adjacent vertices of the 18-gon.  
Then ABC forms the triangle of the problem.

Label the vertices around the 18-gon in sequence beginning at vertex
B and following by vertex C by the letters B, C, d, e, F, G, ...., S.
We use lower case d and e to distinguish between D and E of the 
problem.

Draw segment SG.  Note that arc SG is a 120 degree arc.

The segment OC is the reflection of SG in the diameter BK and 
therefore OC, BK, and SG intersect in a point.  Also, by the half 
angle formula, angle OCB is 50 degrees (because BO is a 100 degree 
arc).  We conclude that OC and SG intersect at D.

The segment Qe is the reflection of SG in the diameter CL and 
therefore Qe, CL, and SG intersect in a point, which we call Z.
Because arc BQ is 60 degrees, ABQ is equilateral, and, similarly, ABe 
is equilateral.  Thus, Qe intersects AB as a perpendicular bisector.
Therefore, Z is equidistant from A and B, that is, ABZ is isosceles.
We conclude that angle ABZ = angle BAZ = 20 degrees, so that angle ZBC
is 80-20 = 60 degrees.  Therefore, Z = E.

Finally, note that BE extends to meet the circle at I since arc CI is
120 degrees.  We compute that angle DEB is half the sum of the arcs SB
and GI, or (20+40)/2 = 30 degrees.

-Doctor Ceeks,  The Math Forum
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Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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