Say $(BLC)$ intersects $AB$ again at $X$. Then the angle conditions imply $AL^2 = AD*AX = AK^2$

Let (BLC)=Γ, AK∩Γ=M, DK∩Γ=N, AL∩Γ=P, DL∩Γ=Q. Then BM∥KN, BP∥LQ. We have \frac{AL}{AP}=\frac{AD}{AB}=\frac{AK}{AM}. ∵AK*AM=AL*AP, ∴AK=AL.

Construct the circumcircle of $LCB$. Then we have that as $\measuredangle ALD = \measuredangle LCB$, we have that $AL$ is tangent to $(LCB)$. Similarly we have that $AK$ is tangent to $(KCB)$. However as $B, K, L, C$ are concyclic, we have that $$\operatorname{Pow}_{(BKLC)}(A) = AL^2 = AK^2 \implies AL = AK, $$as desired. $\blacksquare$

Let $AK\cap (BCLK)=N, AL\cap (BCLK)=M$ Notice that $\angle{AKD}=\angle{BCK}=\angle{BNA}, KD||BN$, similarly, we have $LD||BM$ Then, it is true that $\frac{AD}{AB}=\frac{AK}{AN}=\frac{AL}{AM}$. Since $AK\cdot AN=AL\cdot AM, AK=AL$

Didn't read $K,L$ were inside the triangle but apparently that was enough to solve the problem