Triangle ABC is isoceles, with AB = AC, and the angle at A is
equals to 100°.

Side AB is extended to D, with AD = BC.

Show that the marked angle BCD = 10°.

Choose the point E so that the lines EA and ED both make an
angle of 40° with AD.

Since BC = AD, and angle ABC = angle EAD = 40° and
angle ACB = angle EDA = 40°, the triangles ABC and EAD
are congruent.

Hence AC = AB = EA = ED.

Since AC = AE, triangle ACE is isoceles,
which implies that the angles AEC and ACE must be equal.
Since angle BAC = 100° and angle DAE = 40°, we have
angle EAC = 60°, and so angles AEC and ACE must each equal 60°.

Triangle ACE is therefore equilateral.

It follow that EC = EA, and hence EC = ED, so triangle ECD is
isoceles.

Angle DEC = 100° + 60° = 160°, so angle DCE = 10°,
and angle BCD = 60° - 40° - 10° = 10°