In ΔABC, $$\angle{C}$$ is an obtuse angle. The bisectors of the exterior angles at A and B meet BC and AC produced at D and E respectively. If AB = AD = BE, then $$\angle{ACB}$$ =

From the question,
AB = AD = BE
Let, $$\angle{D}=x$$ and $$\angle{E}=y$$
In$$\triangle{ABD}$$,
$$\angle{ADB}=\angle{ABD}=x$$ ( since AB = AD)
Similarly, in $$\triangle{AEB}$$,
$$\angle{AEB}=\angle{EAB}=y$$
In $$\triangle{ABD}$$,
$$\angle{PAD}=\angle{ADB}+\angle{ABD}= 2x$$ (exterior angle of an triangle)
Since, AD is bisector of $$\angle{A},\angle{PAD}=\angle{CAD}=2x$$
Similarly, BE is bisector of $$\angle{B}$$ so,
$$\angle{QBE}=\angle{CBE}=2y$$
In $$\triangle{ABD}$$,
$$\angle{ADB}+\angle{ABD}+\angle{BAD}=180$$º
⇒ $$x + x +\angle{CAD}+\angle{CAB}=180$$º
⇒ 2x + 2x + y = 180º
⇒ 4x + y = 180º -----eq 1
Similarly, for $$\triangle{ABE}$$,
4y + x = 180º -------eq 2
Solving the two equations simultaneously
x=36º and y=36º
Hence, in $$\triangle{ABC}$$,
$$x + y +\angle{ACB}=180$$º
⇒$$36 + 36 +\angle{ACB}=180$$º
⇒$$\angle{ACB}=108$$º
Hence, option B is correct.