Question

So
(2)
( A B=A C=A D )
( Rightarrow angle mathrm{ABC}=angle mathrm{ACB}=30^{circ} )
( Rightarrow angle B A C=180^{circ}-60^{circ}=120^{circ} )
( mathrm{Now}, angle mathrm{DAC}=180^{circ}-120^{circ}=60^{circ} )
( Rightarrow angle A D C+angle A C D=120^{circ} )
( therefore angle A C D=frac{120^{circ}}{2}=60^{circ} )
( therefore angle mathrm{BCD}=angle mathrm{ACB}+angle mathrm{ACD} )
( =30^{circ}+60^{circ}=90^{circ} )

# Ex.2. ABC is a triangle. The bisectors of the internal angle ZB and external angle ZC intersect at D. If ZBDC= 50°, then ZA is (1) 100° (2) 90° (3) 120° (4) 60°

Solution