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

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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 .


Contributed by: Izidor Hafner (June 2017)
Open content licensed under CC BY-NC-SA




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

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