As shown above, a $\triangle ABC$ is partitioned by $2$ segments $DE$ and $FG$ .

$DE$ is parallel to $BC$ , forming a trapezoid $BCED$ , while $FG$ divides the trapezoid into $2$ kites: the red $BCGF$ and the green $DEGF$ . All side lengths between any points with in these kites are whole integers.

$GF = GC = GE = 17$ , and $BD = 35$ .

What is the area of the blue $\triangle ADE$ ?

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Note
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: Figure not drawn to scale.

The answer is 112.

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Draw diagonals $DG$ and $EF$ intersecting at $H$ , diagonals $CF$ and $BG$ intersecting at $I$ , and a circle centered at $G$ with a radius of $17$ passing through $C$ , $E$ , and $F$ .

Since $CE$ is a diameter of the circle, by Thales' Theorem $\angle CFE = 90°$ . Since all side lengths between any points within the kites are whole integers, $EF$ and $CF$ are integers; therefore, the sides of $\triangle CEF$ are a Pythagorean triple with a hypotenuse of $34$ , which could only be $(16, 30, 34)$ . Therefore, $EF = 16$ and $CF = 30$ .

By properties of diagonals of kites, $\angle FIB = 90°$ and $FI = CI$ ; and since $CF = 30$ , $FI = CI = 15$ . Similarly, $\angle DHF = 90°$ and $FH = HE = 8$ .

$\triangle DHF$ and $\triangle FIB$ are similar by AA similarity, so $\frac{DF}{BF} = \frac{DH}{FI}$ . If $x = DF$ , then $BF = 35 - x$ and by Pythagorean's Theorem on $\triangle DHF$ , $DH = \sqrt{x^2 - 8^2}$ . Now the similarity ratio is $\frac{x}{35 - x} = \frac{\sqrt{x^2 - 8^2}}{15}$ , which has one integer solution of $x = 10$ .

Using Pythagorean's Theorem and the similarity ratio, $DE = 10$ , $DH = 6$ , $HG = 15$ , $GI = 8$ , $BI = 20$ , and $BF = BC = 25$ ; and combining the area of each right triangle gives trapezoid $BCED$ an area of $A_{BCED} = 588$ .

$\triangle ADE$ and $\triangle ABC$ are similar by AA similarity, and has an area ratio of $\frac{DE^2}{BC^2} = \frac{10^2}{25^2} = \frac{4}{25}$ . If $A_{\triangle ADE}$ is the area of $\triangle ADE$ , then $\frac{4}{25} = \frac{A_{\triangle ADE}}{A_{\triangle ADE} + 588}$ , which solves to $A_{\triangle ADE} = \boxed{112}$ .