in any triangle abc prove that bisectors of the interior angle A and the exterior angles at B and C are concurrent i assumed the exterior angles meet at point D, then joined A with D and tried proving AD is angle bisector of angle A. 
But i cant get to the proof. any idea will be helpful
It is enough to consider that a bisector (internal or external) of the angle made by two distinct lines $\ell_1,\ell_2$ meeting at some point $P$ is (part of) the locus of points $Q$ for which $d(Q,\ell_1)=d(Q,\ell_2)$.

If some point $Q$ belongs to the external angle bisector through $B$ and to the external angle bisector through $C$, it fulfills
$$ d(Q,BA) = d(Q,BC) = d(Q,CA) $$
so by $d(Q,BA)=d(Q,CA)$, it belongs to the internal angle bisector through $A$, too.
we construct a parallel line to BC that passes through A and a parallel line to BC that passes through C, we obtain

A parallelogram ABCD.

Consider triangles ABC and ADC

ÐADC º ÐABC.  So, the angle bisectors are the same. We know that the angle bisectors of a triangle are concurrent.  So, the bisectors of ÐABC, ÐDAC, and ÐDCA are concurrent.