Varying foot distance

Let the foot of perpendicular from $A$ to $\overline {BC}$ extended be $D$ and $|\overline {CD}|=x$ . Then $|\overline {AB}|^2=h^2+(2l+x)^2$ and $|\overline {AC}|^2=h^2+x^2$ . Since $\overline {AP}$ is the bisector of $\angle {BAC}$ , therefore $\dfrac{|\overline {AB}|}{|\overline {AC}|}=\dfrac{|\overline {BP}|}{|\overline {PC}|}=\dfrac{l+d}{l-d}$ . Combining these we get $(l-d) ^2(l+x)=d(h^2+x^2)$ . For $d$ to attain maximum, $x$ must equal $\dfrac{(l-d) ^2}{2d}$ . Then $d_{max}=\sqrt {l^2+h^2}-h$ . So $a=1, b=1, c=-1$ and $a+b+c=1+1-1=\boxed 1$