Excircle and parallelogram

Question

Can you provide a proof for the following claim:

Claim. Given any parallelogram $ABCD$ and excircle of triangle $\triangle ABC$ oposite to vertex $A$. An arbitrary tangent $t$ is constructed to the excircle. Let $n_1$ , $n_2$ , $n_3$ , $n_4$ be a signed distances from vertices $A$,$B$,$C$,$D$ to tangent line respectively, such that distances to a tangent from points on opposite sides are opposite in sign, while those from points on the same side have the same sign . Then $n_1+n_3=n_2+n_4$ .

I don't have idea how to start the proof. My first guess is to use barycentric coordinates but I am not sure which triangle to choose as the reference triangle. Any hints are welcomed.

EDIT

As DR SK MOBINUL HAQUE suggested in his comment the more general form of this claim holds as well. You can find a GeoGebra applet that demonstrates his generalization here.

Answer 1

Let the feet of perpendiculars from point $A$, $D$, $B$ and $C$ this line be point $E$, $F$, $G$ and $H$. Extend $BG$ through point $G$ to point $L$ such that $GL=DF$.Also, extend $AE$ through point $E$ to point $M$ such that $EM=CH$. Draw $LM$, $MD$ and $MC$.

Observe that, $EF=GH$. Hence the midpoint of $EH$ and $FG$ are the same.Quadrilaterals $FDGL$ and $MECH$ are parallelograms. $DL$ and $MC$ will pass through this midpoint of $EH$ and $FG$.Also, $MC$ and $DL$ will bisect each other. Thus, quadrilateral $MDCL$ is a parallelogram.

$\Rightarrow ML\parallel CD$

$\Rightarrow ML\parallel AB$

Also, since $AM\parallel LB$, quadrilateral $AMLB$ is a parallelogram.

$\Rightarrow AM=BL$

$\Rightarrow n_{1}+n_{3}=n_{2}+n_{4}$

Answer 2

I will use $'$ to denote the feet of the perpendiculars. So, for example, $A'$ is the foot of the perpendicular from $A$ to $t$. Let the center of the parallelogram be $O$. Recall that the diagonals of a parallelogram bisect each other. Notice that $OO'$ is the common midsegment of trapezoids $AA'C'C$ and $DD'B'B$, completing the proof for the case where the line doesn't cut the parallelogram.

For the case where the line cuts the parallelogram, translate the line high enough so that the theorem proved above can be used. Finally, add the new heights to complete the proof.