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 Check out our web site! http://mathforum.org/dr.math/
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