we need to show $Ar(\Delta APD)=Ar(ABCD)$

Question

$ABCD$ is a quadrilateral. A line through $D$ parallel to $AC$ meets $BC$ produced at $P$ we need to show $$Area(\Delta APD)=Area(ABCD)$$ I tried but did not get properly.

Thank you for helping.

Answer 1

A simple Geogebra screenshot demonstrates the case. In your question you ask to prove, $Ar(APD)= Ar(ABCD)$, which is almost always false. However, as RicardoCruz points out in the comments that you must have meant, $Ar(ABP)=Ar(ABCD)$, which is true, as we can see in the case in the screenshot. To prove it, we can observe that: $$Ar(APB)=Ar(ABC) + Ar(APC)$$ $$Ar(ABCD) = Ar(ABC) + Ar(ADC)$$

[Edit: When $ABCD$ is concave, the $+$ signs above have to replaced by $−$, but it does not contradict our next conclusion.]

Thus, we need to prove that, $Ar(APC)=Ar(ADC)$, which can be done by realizing that quadrilateral $ADPC$ is a trapezium (as we constructed $DP$ and $AC$ parallel). Even if quadrilateral $ABCD$ is concave, $ADPC$ would remain a trapezium, although $PF$ would come inside. And since it is a trapezium, the heights of triangle $ADC$ and $APC$ must be equal (why?), or namely $DE = PF$. So,

$$Ar(ADC)=\frac{1}{2}\cdot AC \cdot DE = \frac{1}{2}\cdot AC \cdot PF = Ar(APC)$$