Solution:

Produce $A P$ to meet $B C$ in $Q$. Join $K E$ and $K F$. Draw perpendiculars from $F$ and $E$ on to $B C$ to meet it in $M$ and $L$ respectively. Let us denote $\angle B K F$ by $\alpha$ and $\angle C K E$ by $\beta$. We show that $\alpha=\beta$ by proving $\tan \alpha=\tan \beta$. This implies that $\angle D K F=\angle D K E$. (See Figure below.)

Since the cevians $A Q$, $B E$ and $C F$ concur, we may write
$$\frac{B Q}{Q C}=\frac{z}{y}, \quad \frac{C E}{E A}=\frac{x}{z}, \quad \frac{A F}{F B}=\frac{y}{x}$$
We observe that
$$\frac{F D}{D E}=\frac{[A F D]}{[A E D]}=\frac{[P F D]}{[P E D]}=\frac{[A F P]}{[A E P]}$$
However, standard computations involving bases give
$$[A F P]=\frac{y}{y+x}[A B P], \quad [A E P]=\frac{z}{z+x}[A C P]$$
and
$$[A B P]=\frac{z}{x+y+z}[A B C], \quad [A C P]=\frac{y}{x+y+z}[A B C]$$
Thus we obtain
$$\frac{F D}{D E}=\frac{x+z}{x+y}$$
On the other hand
$$\tan \alpha=\frac{F M}{K M}=\frac{F B \sin B}{K M}, \quad \tan \beta=\frac{E L}{K L}=\frac{E C \sin C}{K L}$$
Using $F B=\left(\frac{x}{x+y}\right) A B$, $E C=\left(\frac{x}{x+z}\right) A C$ and $A B \sin B=A C \sin C$, we obtain
$$\begin{aligned} \frac{\tan \alpha}{\tan \beta} & =\left(\frac{x+z}{x+y}\right)\left(\frac{K L}{K M}\right) \\ & =\left(\frac{x+z}{x+y}\right)\left(\frac{D E}{F D}\right) \\ & =\left(\frac{x+z}{x+y}\right)\left(\frac{x+y}{x+z}\right)=1 \end{aligned}$$
We conclude that $\alpha=\beta$.