NZMO Round One

Solution:

Assume $T$ lies on $AD$. Then by the angle bisector theorem in triangles $AED$ and $AFD$, we have
$$\frac{AE}{ED} = \frac{AT}{TD} = \frac{AF}{FD}.$$
Applying the angle bisector theorem in triangles $AEB$ and $AFC$, we also have:
$$\frac{AE}{ED} = \frac{AB}{BD}$$
$$\frac{AF}{FD} = \frac{AC}{CD}$$
Combining all of the above, we get that
$$\frac{AB}{BD} = \frac{AC}{CD}, \text{ i.e. } \frac{AB}{AC} = \frac{BD}{CD}. \quad (1)$$
Now extend $AD$ beyond $D$ to meet $BC$ at $A'$. Applying the angle bisector theorem in triangle $ABC$, we have
$$\frac{AB}{AC} = \frac{BA'}{CA'}. \quad (2)$$
Combining (1) and (2) gives
$$\frac{BD}{CD} = \frac{BA'}{CA'}$$
By the converse of the angle bisector theorem, this implies $DA'$ bisects $\angle BDC$, i.e.,
$$\angle BDA' = \angle CDA'.$$
Therefore, by angles on a line, we have
$$\angle BDA = 180^{\circ} - \angle BDA' = 180^{\circ} - \angle CDA' = \angle CDA.$$
Since $\angle BAD = \angle CAD$, $\angle BDA = \angle CDA$, and $AD = AD$, triangles $ABD$ and $ACD$ are congruent (ASA). Then
$$\triangle ABD \equiv \triangle ACD \Longrightarrow AB = AC$$
and we're done.

Solution:
Label the following angles:
$$a = \angle EAT = \angle FAT \qquad \theta_{1} = \angle EDT$$
$$e = \angle AET = \angle DET \qquad \theta_{2} = \angle FDT$$
$$f = \angle AFT = \angle DFT$$
Now consider the product
$$\frac{AT}{ET} \times \frac{ET}{DT} \times \frac{DT}{FT} \times \frac{FT}{AT} = 1.$$
By the law of sines (in triangles $AET$, $EDT$, $DFT$ and $FAT$), this becomes
$$\frac{\sin e}{\sin a} \times \frac{\sin \theta_{1}}{\sin e} \times \frac{\sin f}{\sin \theta_{2}} \times \frac{\sin a}{\sin f} = 1$$
which simplifies to $\sin \theta_{1} = \sin \theta_{2}$. Since $D$ lies inside triangle $ABC$,
$$\theta_{1} + \theta_{2} = \angle EDF = \angle BDC = 180^{\circ} - \underbrace{\angle BCD}_{>0} - \underbrace{\angle CBD}_{>0} < 180^{\circ}.$$
Therefore we have
$$\sin \theta_{1} = \sin \theta_{2} \Longrightarrow \theta_{1} = \theta_{2}.$$
i.e. $\angle EDA = \angle FDA$. We also have $\angle BDF = \angle CDE$ (vertically opposite angles). Combining these,
$$\angle BDA = \angle BDF + \angle FDA$$
$$\qquad = \angle CDE + \angle EDA$$
$$\qquad = \angle CDA.$$
From here we conclude in the same manner as in Solution 1.