Find the area of a quadrilateral in a parallelogram. Assume quadrilateral $ABCD$ is a parallelogram and its area is $S$. And satisfy the following conditions: $AE=BE$, $BF=FC$, $AQ // PC$. Find the area of quadrilateral $APCQ$

• And what have you done? – Yash Jain Feb 8 '18 at 4:17
• @Yash Jain. Thank you for your suggestion. I have used analytic-geometry to calculate a special case: when $ABCD$ is a square. I assume the side length of the square is $a$. then I can calculate the coordinate of points $E$, $P$, $Q$, $D$. then I find $PQ=\frac{1}{5}ED$. So $S_{APCQ} = \frac{1}{5}S_{AECD}$. then $S_{APCQ} = \frac{3}{4} \times S \times \frac{1}{5} = \frac{3S}{20}$ – Laura Feb 8 '18 at 9:27

You have tried a special case and you can get some hint from it. To calculate the area of $APCQ$, we can divide it into two parts: $\triangle{APQ}$ and $\triangle{CPQ}$. Let the intersection point of line $(ED)$ and line$(BC)$ be $T$. By the Menelaus' theorem we have $\frac{AE}{EB} \cdot \frac{BT}{TF} \cdot \frac{FP}{PA}=1$, since $E,F$ are midpoints, we may get that $\frac{FP}{PA}=\frac{3}{2}$. Let the intersection point of the line $(CP)$ and line$(AB)$ be $S$. Apply the Menelaus' theorem again we can get $\frac{AS}{SB} \cdot \frac{BC}{CF} \cdot \frac{FP}{PA}=1$, which implies that $\frac{AS}{SB}=\frac{1}{3}$. Since E is the midpoint of segment $AB$, $S$ is the midpoint of $AE$. As $AQ \parallel CS$, $P$ is the midpoint of $EQ$. The rest is just simple calculation and you can try by yourself. The answer should be the same as what you get in the special case.

• Yes. But you need to change a bunch of Ts to Ss starting halfway through. – almagest Feb 8 '18 at 15:03
• thank you for your suggestions! I have corrected the typos – yifan dai Feb 9 '18 at 1:38

Assume the origin of the coordinate system is $B$. so coordinate of point $B$ is $(0,0)$. And the $x$-axes is $BC$. Assume the coordinate of point $A$ is $(2a,2b)$. The coordinate of point $C$ is $(2c,0)$. Then we can calculate other points' coordinates. $E$ is $(a,b)$, $F$ is $(c,0)$ $D$ is $(2c+2a, 2b)$.

Then the equation of straight line $AF$ is $$y=\frac{-2b}{c-2a}(x-c)$$

the equation of straight line $ED$ is $$y-b=\frac{b}{2c+a}(x-a)$$

so the coordinate of point $P$ is $(\frac{2}{5}(3a+c),\frac{6}{5}b)$

then the equation of straight line $AQ$ is $$y-2b=\frac{3b}{3a-4c}(x-2a)$$

then the coordinate of point $Q$ is $(\frac{4c+7a}{5},\frac{7}{5}b)$

then $$|PQ|=\sqrt{\left(\frac{b}{5}\right)^2+\left(\frac{a}{5}+\frac{2c}{5}\right)^2}$$

and $$|ED|=\sqrt{b^2+(2c+a)^2}$$

then we know $|ED| = 5\times |PQ|$

then $S_{APCQ} = \frac{1}{5}S_{AECD}$ Assume the area of $ABCD$ is $S$. the area of $AECD$ is $\frac{3}{4}S$. So $S_{APCQ}= \frac{3}{4}S \times \frac{1}{5} = \frac{3}{20}S$