This is very trivial and famous . Indeed , we see that $\angle B = 47^{\circ}$ . Notice that $\angle A = 2\angle B$ . LEMMA: If $\angle A = 2\angle B$ , then $a^2=b^2+bc$ .(even converse is also true)/ PROOF : $A=2B \implies A-B=B \implies \sin(A-B)=\sin(B) \implies \sin^2(A)-\sin^2(B) = \sin(B) \cdot \sin(C) \implies a^2=b^2+bc \Box$ . Thus from lemma , we are done. $\Box$

no-no-no..no! it's nice, if and only if without trigonometry. Let a point $B'$ on the line $CA$, such that $A$ - between $C$ and $B'$, $AB'=AB$. So we have that $ \angle AB'B=\angle \frac{BAC}{2}=\angle ABC $ and $CB$ be tangent to the circumcircle of the triangle $ABB'$. Hence $BC^2=CA(CA+AB')=CA(CA+AB')=CA^2+CA\cdot AB$.

INMO 1991!...the values can knock u over though...cunningly set;)