Points $E, F, G, H$ lie on the sides $A B, B C, C D$, and $D A$, respectively, of a square $A B C D$. If $E F G H$ is also a square whose area is $62.5 \%$ of that of $ABCD$ and $CG$ is longer than $EB$ , then the ratio of length of $EB$ to that of $CG$ is

We have a square ABCD with points E, F, G, H on sides AB, BC, CD, DA respectively. These four points form another square EFGH whose area is 62.5% of the original square.

Let's assume the area of square ABCD is 100 units (this makes calculations easier).

Side of ABCD = $\sqrt{100} = 10$ units

Area of EFGH = 62.5 units

Side of EFGH = $\sqrt{62.5} = 2.5\sqrt{10}$ units

---

When we have a square inscribed in another square like this, the four corner triangles (AEH, BFE, CGF, DHG) are congruent.

The triangles are congruent because:

Each triangle has two sides of the original square as its legs

Each triangle has the same angle (90°) at the corner of the original square

The symmetry of the configuration ensures they're identical

This means: $AE = BF = CG = DH$ (let's call this $x$)

And: $EB = FC = GD = AH = 10 - x$

---

Looking at triangle AEH:

$AE = x$ (horizontal leg)

$AH = 10 - x$ (vertical leg)

$EH = \sqrt{62.5}$ (hypotenuse)

Using the Pythagorean theorem:

$AE^2 + AH^2 = EH^2$

$x^2 + (10 - x)^2 = (\sqrt{62.5})^2$

---

$x^2 + (10 - x)^2 = 62.5$

$x^2 + 100 - 20x + x^2 = 62.5$

$2x^2 - 20x + 100 = 62.5$

$2x^2 - 20x + 37.5 = 0$

$4x^2 - 40x + 75 = 0$

Using the quadratic formula:

$x = \frac{40 \pm \sqrt{1600 - 1200}}{8} = \frac{40 \pm \sqrt{400}}{8} = \frac{40 \pm 20}{8}$

This gives us: $x = 7.5$ or $x = 2.5$

---

We have two possible values for $x$:

If $x = 2.5$, then $CG = 2.5$ and $EB = 10 - 2.5 = 7.5$

If $x = 7.5$, then $CG = 7.5$ and $EB = 10 - 7.5 = 2.5$

The problem states that CG is longer than EB.

Therefore: $CG = 7.5$ and $EB = 2.5$

---

The ratio of EB to CG is:

$EB : CG = 2.5 : 7.5 = 1 : 3$

When dealing with inscribed squares, always look for the symmetry that creates congruent triangles. This reduces a complex geometry problem to a simple Pythagorean theorem application!