Hopefully someone can find a better solution... [asy][asy] import olympiad; size(6cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair A = dir(150), C = dir(210), B = dir(330), L = (B+C)/2, K = extension(A,2*foot(L,A,dir(270))-L,B,C), J = 2*foot(L,A,B)-L, T = extension(J,K,A,B); draw(A--B--C--A); draw(K--A--L--J--K); draw(T--L, dashed); draw(anglemark(L,A,B)^^anglemark(C,A,K)); dot("$A$", A, dir(120)); dot("$B$", B, dir(330)); dot("$C$", C, dir(225)); dot("$K$", K, dir(270)); dot("$L$", L, dir(270)); dot("$T$", T, dir(90)); dot("$J$", J, dir(45)); [/asy][/asy] Note that $\overline{AK}$ is a symmedian and the reflection $J$ of $L$ over $\overline{AB}$ lies on $\overline{KT}$ by the angle condition. Let $b = AC$ and $c = AB$ so that $c^2-b^2=1$. The perimeter of $\triangle KTL$ is simply $KL + KJ$. We have $$KL = CL - CK = \frac{1}{2} - 1 \cdot \frac{b^2}{b^2+c^2} = \frac{b^2+c^2-2b^2}{2(b^2+c^2)} = \frac{1}{2(2b^2+1)}.$$By Law of Cosines, \begin{align*} KJ^2 &= KL^2 + JL^2 - 2 \cdot KL \cdot JL \cos \angle KLJ \\ &= \frac{1}{4(2b^2+1)^2} + \left(\frac bc\right)^2 - \frac{1}{2b^2+1} \cdot \frac bc \cdot \frac{-b}{c}\\ &= \frac{1}{4(2b^2+1)^2} + \frac{b^2}{b^2+1} + \frac{b^2}{(b^2+1)(2b^2+1)} \\ &= \frac{1}{4(2b^2+1)^2} + \frac{2b^2(b^2+1)}{(b^2+1)(2b^2+1)} \\ &= \frac{(4b^2+1)^2}{4(2b^2+1)^2} \\ \implies KJ &= \frac{4b^2+1}{2(2b^2+1)}.\end{align*}Therefore, the answer is $$KL + KJ = \frac{4b^2+2}{2(2b^2+1)} = \boxed{1}.$$

Like @above, define $J$ as the reflection of $L$ over $\overline{AB}$ be $J$ and the reflection of $L$ over $\overline{AC}$ as $L'$, Claim 1 : $AKLJ$ is cyclic. Proof : Note that since $\overline{AK}$ is the symmedian, we have $$\angle KLJ = \angle KLA + \angle ALJ = 90^{\circ} - (\angle CAL) + 90^{\circ} - \frac{\angle LAJ}{2} = 180^{\circ} - \angle KAL - \angle CAK -\frac{\angle LAJ}{2}= \angle 180^{\circ} - \angle KAJ \blacksquare$$Thus we have $\angle L'KA = 90^{\circ} - \angle CAK = \angle ALJ = \angle AKJ$. Furthermore, we have $\angle AJK = \angle ALK = \angle AL'K$, hence we have $\Delta AL'K \sim \Delta AJK $. Because they share a same side, they must be congruent. Thus the perimeter of $\Delta KTL$ is just $$ KT+TL+KL = KJ+KL = L'K+KL = LL'=1 $$

since the midpoint of $CH$ is the symmedian point then $AK , AL$ are isogonal hence if $ U,V$ are the symmetrics of $K$ in $AB,AC$ then $AL$ is the bisector of $UV $ thus $LU=LV$. More $KU\cap AB=T$ thus $LU=LT+TK$ but $LV=LC+CV=LB+KC$ therefore $LT+TK+LK=BC$ which ends the proof .

Solution without symmedians: Let $C',L'$ be the reflections of $C,L$ over $AB$ (so $L'$ is what is called $J$ above) and let $M$ be the midpoint of $CH$. It suffices to show that $KL'=KC+L'C'$, in other words, that $KL'$ is a tangent to the circle around $A$ through $C$ and $C'$. But this reduces our problem to $\angle L'AK=\angle L'AM=\alpha$ which is easy to see since $ABC$ and $AC'H$ and hence also the triangle $AL'M$ of midpoints are similar. [asy][asy] import olympiad; size(6cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair A = dir(160), C = dir(200), B = dir(340), L = (B+C)/2, K = extension(A,2*foot(L,A,dir(270))-L,B,C), J = 2*foot(L,A,B)-L, D=2*foot(C,A,B)-C, H=(C+D)/2, M=(C+H)/2 ; draw(A--B--C--A); draw(J--A--D--B); draw(C--D); draw(K--A--L--J--K); draw(arc(A,C,D)); dot("$A$", A, dir(120)); dot("$B$", B, dir(330)); dot("$H$", H, dir(0)); dot("$M$", M, dir(0)); dot("$C$", C, dir(225)); dot("$C'$", D, dir(45)); dot("$K$", K, dir(270)); dot("$L$", L, dir(270)); dot("$L'$", J, dir(45)); [/asy][/asy]

Let $N$ is the reflection of $L$ over $AB$. It is known that $AK$ and $AL$ are isogonal. claim 1: $AKLN$ is cyclic proof: $\angle ANL$ = $\angle ALN$ = $90^{\circ}-\angle BAL$ = $90^{\circ}-\angle KAC$ = $\angle AKC$ claim 2: $AL$ is the angle bisector of $\angle KLT$. proof: $\angle KLA$ = $\angle KNA$ = $\angle ALT$ So $A$ is the cinter of $L-excircle$ of tirangle $KLT$. Let $X$ is the projection of $A$ on $LT$. So $LX$ equal to half perimeter of triangle $KLT$ but $ACLX$ cyclic and $\angle XAL$ = $\angle LAC$ so $XL$ = $CL$ SO the perimeter of triangle $KLT$ is 1.