3. Construct a Triangle Given the Length of an Altitude, the Inradius and the Difference of the Base Angles

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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.



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 .

Proof: .

By construction, the altitude from has length . The circle is the incenter of by steps 1 and 3.

By construction, , so ; that is, .


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




[1] D. S. Modic, Triangles, Constructions, Algebraic Solutions (in Slovenian), Ljubljana: Math Publishers, 2009, pp. 92.

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