Triangle Construction from Angle, Altitude and Median

The applet below illustrates construction of a triangle, given ∠BAC (= α), altitude AH (= h), and median AM (= m).

(Slides are draggable. Use the three on the left to setup the problem, the one on the right to determine the solution.)

Solution

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Copyright © 1996-2018 Alexander Bogomolny

Solution

triangle construction from A, h,M

Start with constructing ΔHAM and a line perpendicular to HM at M. On that line choose an arbitrary point D' and find points B', C' on HM such that ∠MD'B' = ∠MD'C' = 90° - α/2.

Let the circumcircle of ΔB'C'D' meet line AM at point A' on the side of B'. Join A'B' and draw AB||A'B'. B is another vertex of the sought ΔABC.

The solution is explained by the fact that all triangles B'D'M, the circumcircles of ΔB'C'D', point A' are homothetic from M such that all A'B' are parallel to each other and to the "right" one AB, and similarly for B'D'||BD.

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Copyright © 1996-2018 Alexander Bogomolny

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