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

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?

• By good conclusion, I mean, in elements (diagonals, angles or sides) of $ABCD$. – Sawarnik Mar 2 '14 at 16:45

$\triangle ACD$ 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 $\overline{BD}$.

• Took a fast look. I wanted it to be in terms of $ABCD$. See the comment on question? – Sawarnik Mar 3 '14 at 22:26
• @Sawarnik: it would be best to add that to your question, not have it in a comment. – robjohn Mar 3 '14 at 23:33
• Oh, Ok, done. That is the condition which I think gives the question the real difficulty and interest. Btw, I had found one more condition involving $PC$. So I think the problem reduces to finding $PC$ [or $AP$, your answer] in terms of $ABCD$. Could you do it? – Sawarnik Mar 4 '14 at 6:46

$$Area(ABD) = Area (ABCD) = Area (ABP).$$
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).
• Why it is so? I am highly skeptical because playing with GeoGebra leads to me many cases where they are almost equal, that too not in any abnormal situation. I think they are just accuracy errors and somewhere where there lies equalities. Btw, one condition that I found but was not that useful [as I want everything to be a relation in sides or angles of $ABCD$] was: $$PC =\frac{2(ABCD)}{DC\cdot \sin C}$$ – Sawarnik Feb 7 '14 at 10:06