3. Construct a Triangle Given the Length of an Altitude, the Inradius and the Difference of the Base Angles
This Demonstration constructs a triangle given the length of the altitude from to the base , the inradius and the difference of the angles at the base.[more]
Let be a point on a line that will contain the base . Let be on the perpendicular to with . Let be the straight line such that with on the base.
Step 1: Let be a point on such that . Let be the intersection of and the perpendicular to at ; therefore, the distance from to is . Let be the circle with center and radius . The circle is tangent to .
Step 2: Draw a circle with as a diameter.
Step 3: Join to both intersections of and . These two lines are tangent to ; extend them to intersect at and .
Step 4: Draw the triangle .
Let , and .
Theorem: In any triangle , if , then the angle between the altitude from and the angle bisector at is .
By construction, the altitude from has length . The circle is the incenter of by steps 1 and 3.
By construction, , so ; that is, .[less]
 D. S. Modic, Triangles, Constructions, Algebraic Solutions (in Slovenian), Ljubljana: Math Publishers, 2009, pp. 92.