Selection Examinations for the IMO, 2015

Denote the intersection of the lines $CT$ and $AB$ by $U$. We want to prove that
$$\frac{|AU|}{|UB|} = \frac{|AC|}{|CB|}, \qquad (2)$$
since it will follow that $CT$ is the angle bisector of $\angle ACB$.

Since the points $C, E, D$, and $L$ are concyclic, and the lines $AB$ and $CL$ are parallel, we have $\angle AED = \pi - \angle DEC = \angle CLD = \pi - \angle DMA$. Thus the points $A, M, D$, and $E$ are concyclic.
Moreover, we have $\angle EDN = \angle EDA - \angle NDA = \angle EMA - \angle MEB = \pi - \angle BME =$

$\prec MEB = \prec EBN$. Hence the points $N, B, D$, and $E$ are also concyclic.
This gives $\prec NED = \pi - \prec DBN = \pi - \prec DCL = \pi - \prec DEL$. Hence the points $N, E$, and $L$ are collinear.

Since the lines $AB$ and $CL$ are parallel the triangles $MBD$ and $LCD$ have the same angles. Therefore they are similar and we have
$$\frac{|MD|}{|BD|} = \frac{|DL|}{|DC|}. \qquad (3)$$
For the same reason the triangles $NAE$ and $LCE$ also have the same angles. Therefore they are similar and we have
$$\frac{|NE|}{|AE|} = \frac{|EL|}{|EC|}. \qquad (4)$$
Since the lines $AD, BE$, and $CU$ intersect in the same point the Ceva's theorem gives
$$\frac{|AU|}{|UB|} \frac{|BD|}{|DC|} \frac{|CE|}{|EA|} = 1.$$
We also have $\prec ACB = \prec ECD = \prec ELD$ and $\prec DEL = \prec DCL = \prec CBA$. Therefore the triangles $ABC$ and $DEL$ are similar and we get
$$\frac{|LE|}{|DL|} = \frac{|CB|}{|AC|}.$$
Inserting (3) and (4) into the Ceva's theorem equation we get
$$\frac{|AU|}{|UB|} \frac{|LE|}{|DL|} = 1. \qquad (5)$$
Inserting this into (5) we get exactly (2) which we set out to prove. Therefore the line $CT$ is indeed the angle bisector of $\prec ACB$.