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

#### Question

#### Answer

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

(1)

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

DG=\dfrac{1}{2}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

(2)

Addition of (1) and (2) gives

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