26b. Construct a Triangle Given the Length of Its Base, the Angle Opposite the Base and the Length of That Angle's Bisector

Initializing live version
Download to Desktop

Requires a Wolfram Notebook System

Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.

This Demonstration constructs a triangle given the length of its base , the angle at and the length of the angle bisector from to .



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 , .


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




[1] D. S. Modic, Triangles, Constructions, Algebraic Solutions (in Slovenian), Ljubljana: Math Publishers, 2009, p. 57.

Feedback (field required)
Email (field required) Name
Occupation Organization
Note: Your message & contact information may be shared with the author of any specific Demonstration for which you give feedback.