The concurrence of angle bisector, median, and altitude in an acute triangle

$ABC$ is an acute triangle. The angle bisector $AD$, the median $BE$ and the altitude $CF$ are concurrent. Prove that angle $A$ is more than $45$ degrees. Here $D,E,F$ are points on $BC,CA,AB$ respectively.

This question was asked in 4th All-Soviet Union Mathematics Competition in 1970 and the wording of the problem is same.

Solution 1:

From the Trig Ceva theorem it follows that: $$\frac{AF}{FB}\cdot\frac{BD}{DC}=1\tag{1}$$ must hold, or: $$\frac{b \cos A}{a \cos B}\cdot\frac{c}{b}=1,\tag{2}$$

$$\cot A\cdot\frac{\sin C}{\sin(\pi/2-B)}=\cot A\cdot\frac{\sin\widehat{ACB}}{\sin\widehat{FCB}}=1\tag{3}.$$

Since $\widehat{ACB}>\widehat{FCB}$, $\cot A<1$ must hold, so:

$$A>\frac{\pi}{4}.$$

(Thanks to @Blue)