16. Construct a Triangle Given Its Circumradius, Inradius and the Difference of Its Base Angles

This Demonstration shows a construction of a triangle given its circumradius , inradius and the difference of the base angles.
Draw a circle of radius with center and a diameter . (The circle will be the circumcircle of .)
Step 1: Draw a line segment at an angle from to meet at . From , draw a ray at angle from .
Let be on at the Euler distance from , where .
Step 2: Drop a perpendicular from to at and let be on such that .
Step 3: The points and are the intersections of and the perpendicular to at .
By construction, is the circumradius of .
Let , and .
Theorem: Let be the circumcenter of a triangle . If , the angle between the angle bisector at and is .
Proof: The angle subtended over from a point on to the right of is because . So the central angle is twice that: . Triangle is isosceles, so . It follows that
By this theorem and the construction of , is the angle bisector at . Therefore by Euler's triangle formula for , is the incenter of . Since is parallel to and , the distance from to is also and the incircle has center , radius .
By step 1, . The theorem states , so .
The distance is the geometric mean of and , that is of and .


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[1] D. S. Modic, Triangles, Constructions, Algebraic Solutions (in Slovenian), Ljubljana: Math Publishers, 2009 p. 83.
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