the angle of B?

Let the mid point of $\overline {BC}$ be $D$ . Then $B=\angle {ABC}=\dfrac{π}{2}-\dfrac{A}{4}, C=\angle {ACB}=\dfrac{π}{2}-\dfrac{3A}{4}$ , where $A=\angle {BAC}$ . Also, $|\overline {BD}|=|\overline {DC}|, \dfrac{|\overline {AD}|}{\sin B}=\dfrac{|\overline {BD}|}{\sin \dfrac{3A}{4}}\implies \dfrac{|\overline {AD}|}{|\overline {BD}|}=\dfrac{\sin (\dfrac{π}{2}-\dfrac{A}{4})}{\sin \dfrac{3A}{4}}=\dfrac{\cos \dfrac{A}{4}}{\sin \dfrac{3A}{4}}$ ,

$\dfrac{|\overline {AD}|}{|\overline {DC}|}=\dfrac{|\overline {AD}|}{|\overline {BD}|}=\dfrac{\cos \dfrac{3A}{4}}{\sin \dfrac{A}{4}}$ .

From these two we get $\sin \dfrac{3A}{2}=\sin \dfrac{A}{2}\implies A=\dfrac{π}{2}\implies B=\dfrac{3π}{8}=67.5\degree, C=\dfrac{π}{8}=22.5\degree$ or the converse, that is, $B=\dfrac{π}{8}=22.5\degree, C=\dfrac{3π}{8}=67.5\degree$ . So, both $B=\boxed {22.5\degree}$ and $B=\boxed {67.5\degree}$ are valid answers. To satisfy the answer given, the problem should ask for the smallest angle.