In triangle $\bigtriangleup ABC$, the altitude from $A$, the angle bisector of $\angle BAC$, and the median from $A$ to $BC$ divide $\angle BAC$ into four equal angles. What is the measure in degree of $\angle BAC$.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
From the problem statement, its obvious that the isogonal conjugate of the altitude is the median. But that would mean that the median passes through the circumcenter of the triangle. The circumcenter lies on the perpendicular biscector of $BC$. It also lies on the median. So the circumcenter is their intersection point. But that is the midpoint of $BC$. So the circumcenter lies on $BC$, which implies $\angle BAC=90^{\circ}$.
Note: The isogonal conjugate of a cevian is its reflection across the angle biscetor. It is a well known fact that the isogonal conjugate of an altitude passes through the circumcenter. It is also very easy to prove as well!
Hammer with tact.
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