A pre-rmo question

From figure, it is clear that $\angle{ACD}=60$ and $\angle{ABD}=30$

The angle bisector of $\angle{ACD}$ and $\angle{CAD}$ are drawn.

Since angle bisectors are concurrent and intersect at incentre,therfore let $E$ be the incentre of triangle $ACD$ .

Since, $\angle{ACD}= \angle{DBA}=30$ .Therefore, ( $E$ , $A$ , $C$ and $B$ ) are cyclic.

So, $\angle{EAC}$ = $\angle{EBC}$ = $10$ .

from exterior angle theorem

$\angle{AEB}$ =50+20

AGAIN,from exterior angle theorem

$\angle{BCA}$ + $\angle{DBC}$ = 70

BUT, $\angle{EAC}$ = $\angle{DBC}$ =10

So, $\angle{BCA}$ =60

ANSWER= $60-10=\boxed{50}$