ABC is a right triangle, right angled at C and `AB = sqrt2 BC.` Then, find `angle ABC.`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Triangle
We have a right triangle ABC, where angle C is the right angle. We know that \( AB = \sqrt{2} \cdot BC \).

### Step 2: Assign Variables
Let:
- \( BC = x \) (the length of side BC)
- Then, \( AB = \sqrt{2} \cdot x \) (the length of side AB)

### Step 3: Use the Pythagorean Theorem
In a right triangle, the Pythagorean theorem states that:
\[ AB^2 = AC^2 + BC^2 \]
Substituting the known lengths:
\[ (\sqrt{2} \cdot x)^2 = AC^2 + x^2 \]
This simplifies to:
\[ 2x^2 = AC^2 + x^2 \]

### Step 4: Rearranging the Equation
Now, we can rearrange the equation to find \( AC^2 \):
\[ AC^2 = 2x^2 - x^2 \]
\[ AC^2 = x^2 \]
Taking the square root of both sides gives:
\[ AC = x \]

### Step 5: Find the Angles Using Trigonometric Ratios
Now we need to find angle \( ABC \), which we will denote as \( \theta \).

Using the cosine ratio:
\[ \cos(\theta) = \frac{BC}{AB} \]
Substituting the values we have:
\[ \cos(\theta) = \frac{x}{\sqrt{2} \cdot x} \]
This simplifies to:
\[ \cos(\theta) = \frac{1}{\sqrt{2}} \]

### Step 6: Determine the Angle
We know that:
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \]
Thus:
\[ \theta = 45^\circ \]

### Final Answer
The angle \( ABC \) is \( 45^\circ \).

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