**Problem: 100 degree Isosceles
Triangle**

Given an isoscles triangle
ABC with AB = AC and the measure of angle BAC = 100 degrees. Extend
AB to point D such that AD = BC. Now

draw segment CD. What is the size of angle BCD?

Note: There can be several different approaches to this problem. Geometric arguments are preferred to using trigonometry.

Ususally, a solution to a problem
such as this requires some auxilliary constructions.

**Do you want a Hint for one
auxillary construction?**

First, I created the lines of reflection in CD with DE and CE constructed. I constructed point E with the two segments reflected about segments AD and AC.

Second, with D as the center and segment AD as the radius, I constructed arc 1, using E as the center and segment AE as the radius, I constructed arc 2 and I drew AE and DE.

Since we used radius AD as a guide then AD = DE = AE. Thus, triangle ADE is an equilateral triangle.

Therefore, <AGC = 100 degrees because <EAC = 40 degrees and <ACD = 40 degrees, by angles in a triangle measuring 180 degrees.

Because <AGC = 100 degrees, <AGB = 80 degrees by supplementary angles. And <CGE = 80 degrees by opposite angles, just as <BGE = 100 degrees.

Look at segment DC, it is the angle bisector of <ADE, therefore, segment AE is a perpendicular bisector of this angle creating 90 degree angles.

Now, I have <CGE = 80 degrees and <CHE = 90 degrees. And by definition of a triangle being a total of 180 degrees, then <BCD = 10 degrees.

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or compre your solution to the solutions of others to see alternative
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of the 100 degree Isosceles Triangle:**