On square $ABCD,$ points $E,F,G,$ and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34.$ Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P,$ and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$. 


[asy]
size(200);
defaultpen(linewidth(0.8)+fontsize(10.6));
pair A = (0,sqrt(850));
pair B = (0,0);
pair C = (sqrt(850),0);
pair D = (sqrt(850),sqrt(850));
draw(A--B--C--D--cycle);
dotfactor = 3;
dot("$A$",A,dir(135));
dot("$B$",B,dir(215));
dot("$C$",C,dir(305));
dot("$D$",D,dir(45));
pair H = ((2sqrt(850)-sqrt(120))/6,sqrt(850));
pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);
dot("$H$",H,dir(90));
dot("$F$",F,dir(270));
draw(H--F);
pair E = (0,(sqrt(850)-6)/2);
pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);
dot("$E$",E,dir(180));
dot("$G$",G,dir(0));
draw(E--G);
pair P = extension(H,F,E,G);
dot("$P$",P,dir(60));
label("$w$", (H+E)/2,fontsize(15));
label("$x$", (E+F)/2,fontsize(15));
label("$y$", (G+F)/2,fontsize(15));
label("$z$", (H+G)/2,fontsize(15));
label("$w:x:y:z=269:275:405:411$",(sqrt(850)/2,-4.5),fontsize(11));
[/asy]
1. **Define Variables and Ratios:**
   Let the areas of the quadrilaterals \(AEPH\), \(BFPE\), \(CGPF\), and \(DHPG\) be \(w\), \(x\), \(y\), and \(z\) respectively. Given the ratios:
   \[
   w : x : y : z = 269 : 275 : 405 : 411
   \]
   Let \(w = 269a\), \(x = 275a\), \(y = 405a\), and \(z = 411a\). The total area of the square \(ABCD\) is:
   \[
   w + x + y + z = 1360a
   \]

2. **Equal Area Sums:**
   Since \(w + z = x + y = 680a\), it implies that the areas of opposite quadrilaterals sum to the same value. This symmetry suggests that \(AE \cong CG\) and \(DE \cong BG\).

3. **Side Length of the Square:**
   Let the side length of the square \(ABCD\) be \(s\). We need to find \(s\).

4. **Intersection and Symmetry:**
   Let \(I\) be a point on \(AB\) such that \(AI \cong BG\) and \(J\) be a point on \(BC\) such that \(DJ \cong AE\). Let \(O\) be the intersection of \(IJ\) and \(GE\). The quadrilaterals \(AIJD\) and \(BGEA\) are congruent.

5. **Area Calculation:**
   The area of the parallelogram \(HIJF\) is:
   \[
   [HIJF] = [AIJD] - [AHFD] = [BGEA] - [AHFD] = (w + z) - (w + x) = 136a
   \]
   The ratio of the area of \(HIJF\) to the area of the square \(ABCD\) is:
   \[
   \frac{[HIJF]}{[ABCD]} = \frac{136a}{1360a} = \frac{1}{10}
   \]
   Since \(HIJF\) and \(ABCD\) have the same height \(AD\), we get:
   \[
   HI = \frac{1}{10} AB = \frac{s}{10}
   \]

6. **Center of the Square:**
   By symmetry, \(O\) is the center of the square. The areas of the triangles formed by the diagonals are:
   \[
   [AIOE] = \frac{1}{4} \cdot 1360a = 340a
   \]
   The area of \(HIOP\) is:
   \[
   [HIOP] = [AIOE] - [AHPE] = 340a - 269a = 71a
   \]
   Similarly, the area of \(POJF\) is:
   \[
   [POJF] = 65a
   \]

7. **Area of Triangle \(QOP\):**
   Let \(Q\) be a point on \(HF\) such that \(QO \parallel AB\). Since \(O\) is the center of the square:
   \[
   [HIOQ] = [QOJF] = \frac{1}{2}([HIOP] + [POJF]) = 68a
   \]
   Thus, the area of \(QOP\) is:
   \[
   [QOP] = 3a
   \]
   The ratio of the area of \(QOP\) to \(HIJF\) is:
   \[
   \frac{[QOP]}{[HIJF]} = \frac{3a}{71a + 65a} = \frac{3}{136}
   \]
   Since triangle \(QOP\) and parallelogram \(HIJF\) have the same height \(OP\):
   \[
   \frac{QP}{HF} = 2 \cdot \frac{3}{136} = \frac{3}{68}
   \]
   Therefore, \(QP = \frac{3}{2}\).

8. **Finding \(AL\):**
   Let \(K\) be a point on \(AD\) such that \(BK \parallel GE\) and \(L\) be a point on \(BK\) such that \(AL \parallel HF\). We have \(\triangle ALB \sim \triangle QPO\):
   \[
   \frac{AL}{QP} = \frac{AB}{QO}
   \]
   Hence:
   \[
   \frac{AL}{\frac{3}{2}} = \frac{a}{\frac{a}{10}}
   \]
   Thus:
   \[
   AL = 15
   \]

9. **Calculating \(BK\):**
   Since \(BGEK\) is a parallelogram, \(BK = GE = 34\). Let \(BL = b\) and \(LK = 34 - b\). Using the right triangle property:
   \[
   AL^2 = KL \cdot BL
   \]
   Therefore:
   \[
   15^2 = (34 - b) b
   \]
   Solving this quadratic equation:
   \[
   225 = 34b - b^2
   \]
   \[
   b^2 - 34b + 225 = 0
   \]
   Solving for \(b\):
   \[
   b = \frac{34 \pm \sqrt{34^2 - 4 \cdot 225}}{2} = \frac{34 \pm \sqrt{1156 - 900}}{2} = \frac{34 \pm \sqrt{256}}{2} = \frac{34 \pm 16}{2}
   \]
   Thus:
   \[
   b = 25 \quad \text{or} \quad b = 9
   \]
   Since \(BL > LK\), we have \(BL = 25\) and \(LK = 9\).

10. **Area of the Square:**
    The area of the square \(ABCD\) is:
    \[
    AB^2 = AL^2 + BL^2 = 15^2 + 25^2 = 225 + 625 = 850
    \]

The final answer is \(\boxed{850}\).