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

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

Uniteasy Admin · Accepted 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 $$ 	riangle CDF$$ is a special right triangle and its internal angle
$$\angle DCF = 30\degree  $$n
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 $$	riangle DCG $$,
$$\angle DCG<30\degree  $$n
Addition of (1) and (2) gives
$$ \angle ACB = \angle DCF+\angle DCG<60\degree $$

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