If you remember, the Classic Cross Ladders Theorem was used in one of the solutions to this MathCounts problem. The Extended Cross Ladders Theorem, also known as Stengel’s Theorem, applies to cross ladders in a triangle. Specifically, in $\triangle{ABC}$ as shown below, if $BG\parallel DH\parallel FJ\parallel EI$, then $$\dfrac{1}{EI}+\dfrac{1}{DH}=\dfrac{1}{BG}+\dfrac{1}{FJ}$$
It also implies that $$\dfrac{1}{\triangle{AEC}}+\dfrac{1}{\triangle{ADC}}=\dfrac{1}{\triangle{ABC}}+\dfrac{1}{\triangle{AFC}}$$
Question: Let $\triangle{AEF}=3$, $\triangle{ACF}=4$, $\triangle{CDF}=1$. Find the area of quadrilateral $BEFD$.
Solution: Let $x$ be the area of quadrilateral $BEFD$. By Stengel’s Theorem, we have $$\dfrac{1}{3+4}+\dfrac{1}{1+4}=\dfrac{1}{x+4}+\dfrac{1}{4}$$
Solving the above equation, we have $x=\boxed{\dfrac{88}{13}}$.
Challenge: Can you apply Stengel’s Theorem to find the Area of Irregular Pentagon?