# 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
• First, can you check your question, and verify that your vertices are correct? For example, I believe you mean that "OP must have meant to prove that __ABP__=ABCD". If you are interested in the case where ABD=ABCD, then we already know that ABD=ABCD=ABP. Since these are 2 triangles with the same base, hence they must have the same height. It is easy to show that D, P lie on the same side of AC and on the other side of B, hence D,P lie on the same side of AB. Hence, we get that DP is parallel to AB. – Calvin Lin Feb 8 '14 at 23:50
• Oh, I am extremely sorry. I have corrected the question. Perhaps, since the title is contradictory, you should pinged me earlier. Anyways, can you solve the corrected version? Its quite interesting to me. – Sawarnik Feb 9 '14 at 2:52
• If we do not have any idea, a bounty perhaps? – Sawarnik Feb 10 '14 at 9:49