In an acute triangle \triangle ABC, the longest altitude BE is equal to the median CD in length. Show that \angle ACB <60\degree

Collected in the board: Triangle

Steven Zheng posted 1 year ago


Since D is the midpoint of AB, AD = AD

Drop an altitude DF to AC, then DF\parallel BE

DF = \dfrac{1}{2}BE =\dfrac{1}{2}CD

It's found that \triangle CDF is a special right triangle and its internal angle

\angle DCF = 30\degree

Similarly, drop two altitudes from point D and point A to the side of BC. Apparently, DG\parallel AH


Since BE is the longest altitude

DG= < \dfrac{1}{2}BE = \dfrac{1}{2}CD

which shows that in the right triangle \triangle DCG ,

\angle DCG<30\degree

Addition of (1) and (2) gives

\angle ACB = \angle DCF+\angle DCG<60\degree

Steven Zheng posted 1 year ago

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