24c. Construct a Triangle Given the Difference of Base Angles, Length of the Altitude from the Base and the Sum of the Lengths of the Other Two Sides

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This Demonstration constructs a triangle given the difference of the base angles and , the length of the altitude to the base and the sum of the length of the other two sides and .


This problem is solved by constructing a triangle given the length of a side opposite , the length of an angle bisector at and the angle at .


1. Draw a segment .

2. Draw a circle with center and central angle over the chord . Let be the diameter of perpendicular to . Let be the midpoint of .

3. Let be on the extension of so that . Draw the circle with center and radius .

4. Let intersect at and .

5. Draw a circle with center and radius . Let be the intersection of and . Draw the triangle .

6. Let be the intersection of and . Let be the circumcircle of triangle with center .

7. Let be the reflection of across .

Then the triangle satisfies the conditions.


Theorem: the segment is the angle bisector of .

In the isosceles triangle , .

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 , .

Let be the altitude from at . Then . By construction, . So .

Consider triangle . Its exterior angle at is , and interior angles at other vertices are and . So .

Since is symmetric to , . 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.

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