MHB Acute Triangle ABC: Point P and Line Intersections Formula

[anemone]

Here is this week's POTW:

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Consider an acute triangle $ABC$ and let $P$ be an interior point of $ABC$. Suppose the lines $BP$ and $CP$, when produced, meet $AC$ and $AB$ in $E$ and $F$ respectively. Let $D$ be the point where $AP$ intersects the line segment $EF$ and $K$ be the foot of perpendicular from $D$ on to $BC$. Show that $DK$ bisects $\angle EKF$.

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[anemone]

No one answered last week's POTW. (Sadface) However, you can find the suggested solution below:

Spoiler

[TIKZ]\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:C] (C) at (12, 0);
\coordinate[label=above:A] (A) at (4,6);
\coordinate[label=below: Q] (Q) at (9,0);
\coordinate[label=below: M] (M) at (2,0);
\coordinate[label=below: L] (L) at (10,0);
\coordinate[label=below: K] (K) at (7.333333,0);
\coordinate[label=above: F] (F) at (2,3);
\coordinate[label=above: E] (E) at (10,1.5);
\coordinate[label=above: P] (P) at (8,1.2);
\coordinate[label=above: D] (D) at (7.333333,2);
\draw (B) -- (C)-- (A)-- (B);
\draw (B) -- (E);
\draw (A) -- (Q);
\draw (E) -- (F);
\draw (F) -- (C);
\draw (F) -- (M);
\draw (D) -- (K);
\draw (E) -- (L);
\draw [dashed] (F) -- (K);
\draw [dashed] (E) -- (K);
\draw (M) rectangle +(-0.5, 0.5);
\draw (K) rectangle +(-0.5, 0.5);
\draw (L) rectangle +(-0.5, 0.5);
\node (1) at (6.3,0.3) {$\alpha$};
\begin{scope}
\path[clip] (B) -- (K) -- (F) -- cycle;
\draw[thick,blue,double] (K) circle (0.812);
\end{scope}
\node (2) at (8.3,0.3) {$\beta$};
\begin{scope}
\path[clip] (E) -- (K) -- (C) -- cycle;
\draw[thick,blue,double] (K) circle (0.812);
\end{scope}[/TIKZ]Produce $AP$ to meet $BC$ in $Q$. Join $KE$ and $KF$. Draw perpendiculars from $F$ and $E$ to $BC$ to meet it in $M$ and $L$ respectively. Denote $\angle BKF=\alpha$ and $\angle CKE=\beta$. If we can show that $\alpha=\beta$, this implies that $\angle DKF=\angle DKE$.

Since the cevians $AQ,\,BE$ and $CF$ concur, we may write

$\dfrac{BQ}{QC}=\dfrac{z}{y},\,\dfrac{CE}{EA}=\dfrac{x}{z},\,\dfrac{AF}{FB}=\dfrac{y}{x}$

We observe that

$\dfrac{FD}{DE}=\dfrac{[AFD]}{[AED]}=\dfrac{[PFD]}{[PED]}=\dfrac{[AFP]}{[AEP]}$

However, standard computations involving bases give

$[AFP]=\dfrac{y}{y+x}[ABP],\,[AEP]=\dfrac{z}{z+x}[ACP],\,[ABP]=\dfrac{z}{x+y+z}[ABC]$ and $[ACP]=\dfrac{y}{x+y+z}[ABC]$

Thus we obtain $\dfrac{FD}{DE}=\dfrac{x+z}{x+y}$.

On the other hand,

$\tan \alpha=\dfrac{FM}{KM}=\dfrac{FB\sin B}{KM},\,\tan \beta=\dfrac{EL}{KL}=\dfrac{EC\sin C}{KL}$

Using $FB=\left(\dfrac{x}{x+y}\right)AC$ and $AB\sin B=AC\sin C$ we obtain

$\begin{align*}\dfrac{\tan \alpha}{\tan \beta}&=\left(\dfrac{x+z}{x+y}\right)\left(\dfrac{KL}{KM}\right)\\&=\left(\dfrac{x+z}{x+y}\right)\left(\dfrac{DE}{FD}\right)\\&=\left(\dfrac{x+z}{x+y}\right)\left(\dfrac{x+y}{x+z}\right)\\&=1\end{align*}$

$\therefore \alpha=\beta$ and this implies that $\angle DKF=\angle DKE$. In other words, $DK$ bisects $\angle EKF$.