The figure referenced is below:

For the sake of simplicity, assume that the square has sides of length 4. The following reasoning is independent of the actual lengths, and the reason for choosing 4 will become apparent in the explanation.

$\displaystyle Q$ and $\displaystyle X$ are midpoints of their respective sides, so $\displaystyle QS = SX = 2$, making $\displaystyle \overline{QX}$ the hypotenuse of a triangle with legs of length 2 and 2. Therefore,

$\displaystyle (QX ) ^{2}= (SQ)^{2}+ (SX)^{2} = 2^{2}+ 2^{2} = 4+ 4 = 8$.

Also, $\displaystyle QA = 2$, and since $\displaystyle U$ is the midpoint of $\displaystyle \overline{AQ}$, $\displaystyle AU = 1$. $\displaystyle AR = 4$, making $\displaystyle \overline{UR}$ the hypotenuse of a triangle with legs of length 1 and 4. Therefore, 

$\displaystyle (UR)^{2} = (AU)^{2}+ (AR)^{2} = 1^{2}+ 4^{2} = 1+ 16 = 17$

$\displaystyle (UR)^{2} >(QX ) ^{2}$, so $\displaystyle UR > QX$

The altitude of a right triangle from the vertex of its right angle divides the triangle into two smaller triangles each similar to the larger triangle. In particular, 

$\displaystyle \bigtriangleup BXC \sim \bigtriangleup ABC$.

Their corresponding sides are in proportion, so, setting the ratios of the hypotenuses to the short legs equal to each other,

$\displaystyle \frac{AC}{BC} = \frac{BC}{CX}$

$\displaystyle \frac{AC}{13} = \frac{13}{5}$

$\displaystyle \frac{AC}{13} \cdot 13 = \frac{13}{5} \cdot 13$

$\displaystyle AC = \frac{169}{5} = 33 \frac{4}{5 }$

The altitude of a right triangle from the vertex of its right angle divides the triangle into two smaller triangles each similar to the larger triangle. In particular, 

$\displaystyle \bigtriangleup BXC \sim \bigtriangleup ABC$.

Their corresponding sides are in proportion, so, setting the ratios of the long legs to the short legs equal to each other,

$\displaystyle \frac{AB}{BC} = \frac{BX}{XC}$

By the Pythagorean Theorem. 

$\displaystyle (BX)^{2} = (BC)^{2} - (CX)^{2}$

$\displaystyle (BX)^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$

$\displaystyle BX = \sqrt{144} = 12$

The proportion statement becomes

$\displaystyle \frac{AB}{13} = \frac{12}{5}$

$\displaystyle \frac{AB}{13} \cdot 13 = \frac{12}{5} \cdot 13$

$\displaystyle AB = \frac{156}{5}= 31 \frac{1}{5}$

The measure of the angle formed by the two shorter sides of a triangle can be determined to be acute, right, or obtuse by comparing the sum of the squares of those lengths to the square of the length of the opposite side. We compare:

$\displaystyle (AB) ^{2} + (BC)^{2} = 6^{2} + 8 ^{2} = 36 + 64 = 100$

$\displaystyle (AC)^{2} = 12^{2} = 144$

$\displaystyle (AB) ^{2} + (BC)^{2} < (AC) ^{2}$; it follows that $\displaystyle \angle B$ is obtuse, and has measure greater than $\displaystyle 90 ^{\circ }$

The altitude of a right triangle from the vertex of its right angle divides the triangle into two smaller triangles each similar to the larger triangle. In particular, 

$\displaystyle \bigtriangleup BXC \sim \bigtriangleup ABC$.

Their corresponding sides are in proportion, so, setting the ratios of the hypotenuses to the short legs equal to each other,

$\displaystyle \frac{AC}{BC} = \frac{BC}{CX}$

$\displaystyle \frac{AC}{13} = \frac{13}{5}$

$\displaystyle \frac{AC}{13} \cdot 13 = \frac{13}{5} \cdot 13$

$\displaystyle AC = \frac{169}{5} = 33 \frac{4}{5 }$

Each right triangle is a $\displaystyle 45 ^{\circ }-45 ^{\circ }-90 ^{\circ }$ triangle, making each triangle isosceles by the Converse of the Isosceles Triangle Theorem.

Since $\displaystyle \angle B$ and $\displaystyle \angle E$ are the right triangles, the legs are $\displaystyle \overline{AB}, \overline{BC}, \overline{DE}, \overline{EF}$, and the hypotenuses are $\displaystyle \overline{AC}, \overline{DF}$.

By the $\displaystyle 45 ^{\circ }-45 ^{\circ }-90 ^{\circ }$ Theorem, $\displaystyle AB \sqrt{2} = AC$ and $\displaystyle EF \sqrt{2} = DF$.

$\displaystyle AC > DF$, so $\displaystyle AB \sqrt{2} > EF \sqrt{2}$ and subsequently, $\displaystyle AB > EF$.

The altitude of a right triangle from the vertex of its right angle divides the triangle into two smaller, similar triangles. 

The similarity ratio of $\displaystyle \bigtriangleup AXB$ to $\displaystyle \bigtriangleup BXC$ can be found by determining the ratio of one pair of corresponding sides; we will use the short leg of each, $\displaystyle \overline{XB}$ and $\displaystyle \overline{XC}$.

$\displaystyle \overline{XB}$ is also the long leg of $\displaystyle \bigtriangleup BXC$, so its length can be found using the Pythagorean Theorem:

$\displaystyle (BX)^{2} = (BC)^{2} - (CX)^{2}$

$\displaystyle (BX)^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$

$\displaystyle BX = \sqrt{144} = 12$

The similarity ratio is therefore

$\displaystyle \frac{XB}{XC} = \frac{12}{5}$.

The ratio of the areas is the square of this ratio:

$\displaystyle \left ( \frac{12}{5} \right )^{2} = \frac{12^{2}}{5^{2}} = \frac{144}{25}$ - that is, 144 to 25.

By the Pythagorean Theorem,

$\displaystyle x^{2 } + 12 ^{2 } = 18 ^{2 }$

$\displaystyle x^{2 } + 144 = 324$

$\displaystyle x^{2 } + 144 -144 = 324-144$

$\displaystyle x^{2 } = 180$

$\displaystyle x = \sqrt{180} = \sqrt{36} \cdot \sqrt{5} = 6 \sqrt{5}$

$\displaystyle m \angle B = 90^{\circ}$

$\displaystyle m \angle C = 45 ^{\circ }$

The sum of the measures of the angles of a triangle is , so:

$\displaystyle m \angle A + m \angle B + m \angle C = 180$

$\displaystyle m \angle A +90 + 45 = 180$

$\displaystyle m \angle A +135= 180$

$\displaystyle m \angle A +135-135= 180-135$

$\displaystyle m \angle A = 45^{\circ }$

This is a $\displaystyle 45^{\circ }-45^{\circ }-90^{\circ }$ triangle, so its legs $\displaystyle \overline{AB}$ and $\displaystyle \overline{BC}$ are congruent. The quantities are equal.

The area of a triangle is half the product of its height and its base; in a right triangle, the legs, being perpendicular, can serve as these quantites.

The triangle in the diagram has area 

$\displaystyle \frac{1}{2} \times 30 \times 18 = 270$ square inches.

An isosceles right triangle has two legs of the same length, which we will call $\displaystyle L$. The area of that triangle, which is the same as that of the one in the diagram, is therefore 

$\displaystyle \frac{1}{2} L ^{2} = 270$

$\displaystyle \frac{1}{2} L ^{2}\cdot 2 = 270 \cdot 2$

$\displaystyle L ^{2} = 540$

$\displaystyle L = \sqrt{540}$

$\displaystyle L = \sqrt{36}\cdot \sqrt{15} = 6 \sqrt{15}$ inches.