Prove that (KO)/(KL) = (1)/(sqrt(2)) given that KOP ||| KLM and Area( KOP) = Area(quadrilateral OLMP).

Given that $OP \parallel LM$.

$$\angle K = \angle K \quad (\text{Common angle})$$ $$\angle KOP = \angle KLM \quad (\text{Corresponding angles, since } OP \parallel LM)$$ $$\angle KPO = \angle KML \quad (\text{Corresponding angles, since } OP \parallel LM)$$

Therefore, by Angle-Angle-Angle (AAA) similarity criterion:

$$\boxed{\triangle KOP ||| \triangle KLM}$$

Since $\triangle KOP ||| \triangle KLM$ (from Step 1), the ratio of corresponding sides is equal to the ratio of corresponding altitudes.
$KY$ is the altitude from $K$ to $OP$ in $\triangle KOP$.
$KX$ is the altitude from $K$ to $LM$ in $\triangle KLM$.
$OP$ and $LM$ are corresponding bases.
Thus, the ratio of the altitudes is equal to the ratio of the corresponding sides:

$$\boxed{\frac{KY}{KX} = \frac{OP}{LM}}$$

Let $A_1 = \text{Area}(\triangle KOP)$.
Let $A_2 = \text{Area}(\triangle KLM)$.
We are given that $\text{Area}(\triangle KOP) = \text{Area}(\text{quadrilateral } OLMP)$.
So, $A_1 = \text{Area}(\text{quadrilateral } OLMP)$.
The area of $\triangle KLM$ is the sum of the area of $\triangle KOP$ and the area of quadrilateral $OLMP$:

$$A_2 = \text{Area}(\triangle KOP) + \text{Area}(\text{quadrilateral } OLMP)$$

Substitute the given information:

$$A_2 = A_1 + A_1 = 2A_1$$

Now, find the ratio of the areas of the two similar triangles:

$$\frac{\text{Area}(\triangle KOP)}{\text{Area}(\triangle KLM)} = \frac{A_1}{A_2} = \frac{A_1}{2A_1} = \frac{1}{2}$$

For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides:

$$\frac{\text{Area}(\triangle KOP)}{\text{Area}(\triangle KLM)} = \left(\frac{KO}{KL}\right)^2$$

Substitute the ratio of areas:

$$\frac{1}{2} = \left(\frac{KO}{KL}\right)^2$$

Take the square root of both sides:

$$\frac{KO}{KL} = \sqrt{\frac{1}{2}}$$ $$\frac{KO}{KL} = \frac{\sqrt{1}}{\sqrt{2}}$$ $$\boxed{\frac{KO}{KL} = \frac{1}{\sqrt{2}}}$$