Solving math problems in senior three.

Because CD is the bisector of ∠C, ∠ACD=∠BCD.

Because BD=DC, ∠DCB=∠DBC.

So ∠ACD=∠DCB=∠CBD

And because ∠ A+∠ ACD+∠ DCB+∠ CBD =180 = ∠ A+3x, X = 15.

Drawing: make the extension line of AB, so that the extension line of AB is perpendicular to point C ... and set the vertical foot as E.

Because ∠ A = 135, ∠ EAC = ∠ ACE = 45.

So ∠ DCE = 60 and because DC= radical number 3, EC=cos60* radical number 3= radical number 3/2.

At this point, it's simple. Select a solution. I choose ∠ EBC = 15.

Sin 15 = EC/BC, so BC= (root number 3/2) 3/2)/sin 15.

Because sin15 = sin (45-30) = sin45cos30-cos45sin30 = (61/21/2)/4 = ()/4.

So BC= (root number 3 /2) * 4/(square root of 6-root number 2) = (square root of root number 3 2+6)/2.

I don't know if there is anything wrong. I haven't done it for a long time. For reference only.

You can also make point D perpendicular to BC.

Cos 15 =( 1/2bc)/ radical 3 BC= radical 3 * cos15 * 2 = (the square root of 3 radicals 2+6) /2.