Constructing an Equal-Area Triangle from a Quadrilateral

 Claim. Let $ABCD$ be a convex quadrilateral such that $AB \not\parallel CD$. Through $B$, draw the line parallel to $CD$, and through $C$, draw the line parallel to $AB$; let these two lines meet at $O$. Let $M$ and $N$ be the midpoints of $BO$ and $CO$, respectively. Let $A'$ be the reflection of $A$ about $N$, and let $D'$ be the reflection of $D$ about $M$. Then

$$[ABCD]=[A'D'O].$$

 Proof. Let the diagonals $AC$ and $BD$ meet at $P$. Since $N$ is the midpoint of both $CO$ and $AA'$, the half-turn about $N$ sends $A$ to $A'$ and $C$ to $O$. Hence

$$A'O=AC \qquad \text{and} \qquad A'O \parallel AC.$$

Similarly, since $M$ is the midpoint of both $BO$ and $DD'$, the half-turn about $M$ sends $D$ to $D'$ and $B$ to $O$. Therefore

$$D'O=BD \qquad \text{and} \qquad D'O \parallel BD.$$

Let

$$\theta=\angle BPC,$$

the angle between the diagonals $BD$ and $AC$. Since $A'O \parallel AC$ and

$D'O \parallel BD$, we have

$$\sin \angle A'OD'=\sin \theta.$$

Thus

$$[A'D'O]=\frac12 \cdot A'O \cdot D'O \cdot \sin \angle A'OD'=\frac12 \cdot AC \cdot BD \cdot \sin \theta.$$

On the other hand, the area of a convex quadrilateral is equal to one half the product of its diagonals times the sine of the angle between them. Hence

$$[ABCD]=\frac12 \cdot AC \cdot BD \cdot \sin \theta.$$

Therefore,

$$[A'D'O]=[ABCD],$$

as desired.