When is $Ar(APD)=Ar(ABCD)$?

Question

This question arose while I was answering this question, (we need to show $Ar(\Delta APD)=Ar(ABCD)$). First the original question:

$ABCD$ is a quadrilateral. A line through $D$ parallel to $AC$ meets $BC$ produced at $P$ we need to show $$Area(\Delta APD)=Area(ABCD)$$

Its easy to see that the OP must have meant to prove, $Area(\Delta ABP)=Area(ABCD)$, the proof of which is given in my answer. However, it set me thinking when actually $Area(\Delta ADP)=Area(ABCD)$ is true? I have not been able to derive any good conclusion (I mean, in terms of the elements (diagonals, angles or sides) of $ABCD$). Can anyone help?

Answer 1

$\triangle ACD$ has the same area as $\triangle ACP$ since they both have the same base and altitude. Thus, $ABCD$ has the same area as $\triangle ABP$. Thus, we need to find when the area of $\triangle ABP$ has the same area as $\triangle APD$. This happens precisely when the distance from $D$ to $\overline{AP}$ is the same as the distance from $B$ to $\overline{AP}$ so that the triangles have the same altitude (they already have the same base). This happens when $\overline{AP}$ bisects $\overline{BD}$.

Thus, the area of $ABCD$ is the same as the area of $\triangle APD$ precisely when $\overline{AP}$ bisects $\overline{BD}$.

Answer 2

Your question reduces to asking when is

$$Area(ABD) = Area (ABCD) = Area (ABP).$$

This happens if and only if $ AB \parallel DP$. But since we are given that $DP \parallel AC$, such a situation will not arise (except in the degenerate case).