, 2022

Proof 1: If $AB \parallel CD$ and $AD \parallel BC$, then $ABCD$ is a parallelogram. In this case, $P$ is the midpoint of $AC$ and $AE \parallel CF$. So $PE = PF$.

We may assume that $AB$ is not parallel to $CD$ from now on. We first show that $AB + AD = CB + CD$. As shown in the following picture, we may assume that the extends of $BA$ and of $CD$ meet at point $T$. Let $\odot K$ be the excircle of $\triangle TBC$ in $\angle B$; meeting the lines $AB$, $BC$, $CD$ at $X$, $Y$, $Z$, respectively.

Since the internal homothetic center of $\odot I$ and $\odot J$ lies on the line $IJ$, and since $AC$ is an inner tangent of the two circles, we deduce that the inner homothetic center of $\odot I$ and $\odot J$ is $P$.
It is well known that the internal homothetic center $P$ of $\odot I$ and $\odot J$, the external homothetic center $B$ of $\odot I$ and $\odot K$, and the internal homothetic center of $\odot J$ and $\odot K$ are colinear. So the internal homothetic center of $\odot J$ and $\odot K$ lies on the line $BP$. Since $CD$ is an internal tangent line of $\odot J$ and $\odot K$, the internal homothetic center of $\odot J$ and $\odot K$ is precisely point $D$. Moreover, $AD$ is tangent to $\odot K$, say at point $W$.
From this, we deduce that
$$\begin{aligned} AB + AD &= (BX - AX) + (AW - DW) = BX - DW \\ &= BY - DZ = (CB + CY) - (CZ - CD) = CB + CD. \end{aligned}$$
In other words,
$$AB + AD = CB + CD. \tag{1}$$
Let $\odot U$ and $\odot V$ denote the excircles of $\triangle ABD$ in $\angle ABD$ and in $\angle ADB$, respectively. Let $\odot Q$ and $\odot R$ denote the excircles of $\triangle BCD$ in $\angle CBD$ and in $\angle BDC$, respectively.
Now, we prove that $UR$ and $VQ$ meet at $P$. Consider $\odot K$, $\odot U$, and $\odot R$. It is well known that the external homothetic center $A$ of $\odot K$ and $\odot U$, the internal homothetic center $C$ of $\odot K$ and $\odot R$, and the internal homothetic center of $\odot U$ and $\odot R$ are colinear. Yet $BD$ is the internal tangent line of $\odot U$ and $\odot R$; so $P$ is the internal homothetic center of $\odot U$ and $\odot R$. It follows that $UR$ passes through $P$. By a similar argument, $VQ$ passes through $P$.

Next, we prove that $UQ \perp BD$ and $VR \perp BD$. Extend $BD$ to intersect $\odot U$ and $\odot Q$ at $L$ and $L'$, respectively. Then
$$BL = \frac{1}{2}(AB + AD + BD), \quad BL' = \frac{1}{2}(CB + CD + BD).$$
Combining this with (1), we deduce that $BL = BL'$, i.e. $L$ and $L'$ coincide. So $UQ \perp BD$. Similarly, $VR \perp BD$.
Since $EF \perp BD$, So $UQ \parallel EF \parallel VR$. Thus,
$$\frac{PE}{UQ} = \frac{VP}{VQ} = \frac{RF}{RQ} = \frac{PF}{UQ},$$
So $PE = PF$. We are done.