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

Question

$ABC$ 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.

Answer 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)

Answer 2

A coordinate proof: $A=(0,0),F=(1,0),C=(1,t).$ Then $E=(1/2,t/2),$ and let the concurrency point be $P$, and let $h=AC=\sqrt{1+t^2}.$ From $AC/AF=PC/PF,$ a fact about angle bisectors, we can calculate $P=(1,c/(1+h)).$ Finally $B$ is on the $x$ axis where line $EP$ crosses it, and we find $B=(h/(h-1),0).$ We are to show the claim: if $t \le 1$ then the triangle is obtuse, i.e. the angle at $C$ is more than 90. [This is the contrapositive to saying: If the triangle is acute, then the angle at $A$ is at least 45.]

Now start by noting that for $t=1$ we get $B=(2+\sqrt{2},0)$ so for $t=1$ so we get an obtuse triangle [angle at $C$ is >90] since in this case triangle $AFC$ is an equilateral right triangle but triangle $BFC$ has $BF>FC.$

Next note that as $t$ is decreased from $1$ towards $0$ the angle $ACF$ is increasing, and also, since the length $BF$ is also increasing (the derivative of $h/(h-1)$ is negative and we're decreasing the parameter $t$), the angle $FCB$ is increasing. Combining these we find that the angle $ACB$, which was already obtuse in the $t=1$ case, becomes even more obtuse as $t$ decreases toward $0$.

Alternate (non calculus) finish to argument: Assuming angle $A=CAF$ is $\le 45,$ we have both angle $ACF \ge 45,$ as well as $h=AC\le\sqrt{2}.$ Also, the length $FB$ is strictly greater than $1$ if and only if $h/(h-1) > 2.$ Using just algebra, $h/(h-1) > 2$ is equivalent to $h < 2,$ so that since in fact we have $h \le \sqrt{2}$ it follows that $h <2 $ to finish the argument, since we have the angle $FCB$ strictly greater than 45, which with the other angle $FCA$ being at least 45 makes the angle $ACB$ more than 90, and the triangle is obtuse rather than acute.