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 Skip Navigation LinksMath Help > Geometry > Polygons and Triangles > Right Triangle Median, Altitude, Bisector

The median CM and altitude CH of triangle ABC are reflections about the bisector of angle C iff ABC is a right triangle.

Proof:

Suppose triangle ABC is a right triangle, shown inscribed in a circle with diameter AB and center M.  The bisector of angle ACB is shown here in purple, along with the perpendicular bisector of AB.  The bisector of angle ACB also bisects angle MCH because the purple angle inscribed in the circle is half that of the central angle, M, because they are both subtended by the same arc.

Thus, when ABC is a right triangle, the median CM is the reflection about the right angle's bisector of the altitude CH.

Now, let ABC be a triangle such that the bisector of angle ACB also bisects angle MCH.  Draw a circle whose diameter is AB, and whose center is M.  In that circle, inscribe an angle such that one of its rays is the perpendicular bisector of AB, and the other passes through C, and intersects the circle at C'.  Note that central angle M shares one ray with the inscribed angle, and the other passes through C, intersecting the circle at C''.  Note, too, that the measure of the central angle is exactly double the measure of the inscribed angle, so C' = C''.  Now two different lines intersect at C, and also at C', which means C = C', and is a point on the circle, which means ABC is a right triangle.

Internet references

Related pages in this website:

Summary of geometrical theorems

Law of Sines - Given triangle ABC with opposite sides a, b, and c, a/(sin A) = b/(sin B) = c/(sin C) = the diameter of the circumscribed circle.

Circumscribed Circle - The radius of a circle circumscribed around a triangle is R = abc/(4K), where K is the area of the triangle.

Inscribed Angle -- proof that an angle inscribed in a circle is half the central angle that is subtended by the same arc

Triangle Trisection -- If a point, P, on the median of triangle ABC is the isogonal conjugate of point Q, on the altitude of ABC, then ABC is a right triangle.

Isogonal Conjugates -- the relationship between the orthocenter and circumcenter of a triangle

Ceva's Theorem -- In triangle ABC, arbitrary cevians AD, BE and CF, meet at a point P iff AF/FB � BD/DC � CE/EA = 1

 

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