Let $P$ be an interior point of an acute-angled triangle $ABC$. The line $BP$ meets the line $AC$ at $E$, and the line $CP$ meets the line $AB$ at $F$. The lines $AP$ and $EF$ intersect each other at $D$. Let $K$ be the foot of the perpendicular from the point $D$ to the line $BC$. Show that the line $KD$ bisects the angle $\angle{EKF}$.
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Attempt
Extend $EF$ to intersect line $BC$ at $G$. Note that the pencil formed by the complete quadrangle $BCEF$ of the four lines based at each intersection of opposite sides is harmonic. Thus the cross-ratio $(G,D,F,E) = -1$. This with the fact that $\angle{DKG} = 90^{\circ}$ should prove that $KD$ is a bisector, but I am not sure how.
$\pi_1$ is the circle that circum-scribes KEF. It cuts KD produced at H and BC at S. By converse of Thales, HS is a diameter of that circle. If US is a tangent to $\pi_1$, then, $\angle TSU = 90^0$.
Let HS cuts EF at T.
Claim:- $\pi_2$, the red dotted circle, that circum-scribes SKD also passes through T.
Proof:-
(1) SD is a diameter of $\pi_2$ for the same reason.
(2) Let $\pi_2$ cuts US at V such that $\angle TSV = \angle TSU = 90^0$. Note that V and S are points on the circumference of $\pi_2$.
The above implies T is also on $\pi_2$ and therefore, $HS \bot FE $ at $T$.
In fact, $HS$ is the perpendicular bisector of $FE$ because $EF$ is a chord perpendicular to the diameter $HS$.
Then, $\alpha = \alpha_1 = \beta_1 = \beta$.