jmc

Let's sketch our triangle first. We can see that the shortest angle bisector will be from the vertex $A$ (we leave the proof to the reader). We will call that bisector $AD.$ Since $\angle BAD=\angle CAD$ by definition and $\angle ABC=\angle ACB$ since $△ABC$ is isosceles, we can see that $\angle ADB=\angle ADC=90^{\circ }.$ This is useful, since we know that $AC=14$ and $DC=\frac{1}{2}\cdot BC=13.$ Therefore, we use the Pythagorean theorem to find $A{D}^2=A{C}^2-C{D}^2=14^2-13^2=27.$ Therefore, our answer is $AD=\boxed{3\sqrt{3}}.$