5. Construct a Triangle Given Its Circumradius, the Difference of Base Angles and the Sum of the Other Two Sides

This Demonstration constructs a triangle given the circumradius , the difference of angles at the base and the sum of the lengths of the sides and .
Draw a (blue) circle with center , radius and a diameter . Let be a point on such that .
Step 1: Draw a (red) circle with diameter and center , the midpoint of . Let be on such that .
Step 2: Let intersect at .
Step 3: Let be symmetric to across .
Step 4: Draw the triangle .
Let and .
Let be symmetric to across . Let be the intersection of and .
The angle between the angle bisector from and the altitude from is ; also, .
So is the bisector of the angle at . Since , and , , so .


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[1] D. S. Modic, Triangles, Constructions, Algebraic Solutions (in Slovenian), Ljubljana: Math Publishers, 2009, pp. 96.
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