A geometry problem by SHANKH SURI

$B$ lies on the internal bisector of $\angle ADC$ . The external angle to $\angle BAD$ is $180^\circ - 80^\circ - 20^\circ = 80^\circ$ , hence $BA$ is the external bisector of $\angle CAD$ . As $B$ also lies on the internal bisector of $\angle ADC$ , it follows that $B$ is the $D$ -excenter of $\triangle ADC$ , hence $BC$ is the external angle bisector of $\angle ADC$ . It's easy to find $\angle ACD= 60^\circ$ , hence $\angle BCA = 60^\circ$ , and $\angle DBC = 10^\circ$ .

THIS IS A QUESTION BASED ON X-CENTRE. WE FIND OUT THAT BD IS ANGLE BISECTOR OF CDA. NOW PRODUCE CA TOWARDS "A" VERTEX UP TO K NOW DAK = 120. NOW PRODUCE AB TOWARDS "A" UP TO L NOW BL IS THE ANGLE BISECTOR OF DAK. FURTHER BY THE PROPERTY OF AN X-CENTRE THAT IT DOES EXIST IN EVERY TRIANGLE WE CONCLUDE BC IS ALSO AN ANGLE BISECTOR.EXTEND BC TOWARDS C UP TO M AND AC PRODUCED TO TO N TOWARDS C . NOW MCN = 60 = ACB. FURTHER IN TRIANGLE ABC 80 + 60 + 30 + CBD = 180 CBD = 10