A Described Quandary of an Inscribed Quadrilateral

Since it is given that $\angle E = \angle F$ , and $\angle DAE = \angle BAF$ by vertical angles, $\triangle ABF \sim \triangle ADE$ by AA similarity.

Since $\angle ADC + \angle ABC = 180°$ as opposite angles in an inscribed quadrilateral and $\angle ABF + \angle ABC = 180°$ as a straight line, $\angle ADC = \angle ABF$ , and since $\angle F = \angle F$ by the reflexive property, $\triangle ABF \sim \triangle CDF$ by AA similarity. Likewise, $\triangle AED \sim \triangle CBE$ . Therefore, all four triangles $\triangle ABF$ , $\triangle ADE$ , $\triangle CDF$ , and $\triangle CBE$ are similar to each other.

Since $\angle ABF + \angle ABC = 180°$ from a straight line and $\angle ABF = \angle ABC$ from similar triangles, $\angle ABF = \angle ABC = 90°$ . By a similar argument, $\angle CDF = \angle ADE = 90°$ .

Since the ratio of the perimeters equal the ratio of the sides in similar triangles, $\frac{P_{\triangle ADE}}{P_{\triangle ABF}} = \frac{AD}{AB}$ , or $\frac{12}{28} = \frac{AD}{7}$ , which solves to $AD = 3$ .

By Pythagorean's Theorem on $\triangle ADE$ , $3^2 + DE^2 = AE^2$ , and by the perimeter of $\triangle ADE$ , $3 + DE + AE = 12$ , and these two equations solve to $DE = 4$ and $AE = 5$ . The area of $\triangle ADE$ is then $A_{\triangle ADE} = \frac{1}{2} \cdot AD \cdot DE = \frac{1}{2} \cdot 3 \cdot 4 = 6$ .

The scale factor $k$ of similar triangles $\triangle CBE$ to $\triangle ADE$ is $k = \frac{BE}{DE} = \frac{5 + 7}{4} = 3$ , so the area of $\triangle CBE$ is $A_{\triangle CBE} = k^2 \cdot A_{\triangle ADE} = 3^2 \cdot 6 = 54$ .

Therefore, the area of quadrilateral $ABCD$ is $A_{ABCD} = A_{\triangle CBE} - A_{\triangle ADE} = 54 - 6 = \boxed{48}$ .

Let $(DA, AB, BC, CD) = (a, b, c, d)$ . Then

$AF = b \dfrac{ab+cd}{d^2-b^2}$

$BF = b \dfrac{cb+ad}{d^2-b^2}$

$AE = a \dfrac{da+bc}{c ^2-a^2}$

$DE = a \dfrac{ba+dc}{c^2-a^2}$

We have

$b = 7$
$a AF = b DE$
$b + AF + BF = 28$
$a + AE + DE = 12$

Solve for $a, b, c, d$ to get

$(a,b,c,d) =(3,7,9,11)$

Letting $s = \dfrac{1}{2}(a+b+c+d)$ , we find the area

$A = \sqrt{(s-a)(s-b)(s-c)(s-d)} = 48$

Use chord theorem and similarity of two triangles to solve for remaining two sides of quadrilateral. The side d=7(12/28)=3, as two triangles are similar.

The first integer area solution for quadrilateral was for sides 7,9,11,3.

There are general solutions, 7, b>0, (2/3)b+5 and 3. The solution for all integer sides including two triangles is for b=27, but area is ~114.2.

Answer=48