26b. Construct a Triangle Given the Length of Its Base, the Angle Opposite the Base and the Length of That Angle's Bisector
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This Demonstration constructs a triangle given the length of its base , the angle at and the length of the angle bisector from to .[more]
1. Draw the segment of length . Draw a circle with center and central angle over the chord .
2. Let be the diameter of perpendicular to . Let be the midpoint of .
3. Let the point be on the extension of the segment such that . Draw a circle with center and radius .
4. Draw the line through and that intersects at the points and .
5. Draw the circle with center and radius . The point is one of the points of intersection of and . Draw triangle .
Theorem: The segment is the angle bisector of .
In the isosceles triangle , and . Since , . So is on . Therefore is a tangent of and the power of with respect to the circle is .
The power of with respect to is . By construction . So , .[less]
Contributed by: Izidor Hafner (October 2017)
Open content licensed under CC BY-NC-SA
 D. S. Modic, Triangles, Constructions, Algebraic Solutions (in Slovenian), Ljubljana: Math Publishers, 2009, p. 57.