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