Special case of the Lemoine point (intersection of symmedians) A triangle $ABC$, whose symmedians intersect at $L$, has an altitude $BT$ drawn. 
Given that $\widehat{TLA} = 2\cdot\widehat{TBA}$,  I have to prove that $\widehat{TLC} = 2\cdot\widehat{TBC}$.
Any hints are welcome.
 A: Let we rename $T$ as $H_B$.
The midpoint $M_C$ of $AB$ is the center of the circumcircle of $H_B AB$, in particular $2\cdot\widehat{H_B BA}=\widehat{AM_C H_B}$ and the given relation implies that the symmedian point $L$ lies on the circumcircle of $AM_C H_B$. It is not difficult to prove$^{(*)}$ that in general the symmedian through $B$ always goes through the intersection of the circumcircles of $AH_B M_C$ and $CH_B M_A$. It follows that the given hyphotesis imply that $L$ lies on the circumcircle of $CH_B M_A$ too, so $\widehat{H_B L C}=2\cdot \widehat{H_B B C}$ is granted.

(*) Assume that $J$ is the depicted (but unnamed) intersection of the circumcircles of $AH_B M_C$ and $CH_B M_A$. $J$ "sees" $M_C H_B$ under an angle equal to $\pi-\widehat{A}$ and $M_A H_B$ under an angle equal to $\pi-\widehat{C}$, hence $B M_C J M_A$ is a cyclic quadrilateral and 
$$ \frac{d(J,BA)}{d(J,BC)}=\frac{JM_C}{JM_A}=\frac{M_C H_B}{M_A H_B} =\frac{AM_C}{CM_A}=\frac{AB}{CB} $$
by the sine theorem applied to $JH_B M_C$ and $JH_B M_A$. The last identity proved that $J$ lies on the symmedian through $B$.
