`DeltaABC ` is isosceles with AC = BC . If `AB^(2)= 2AC^(2)` then find the measure of `angleC `

Answer

To solve the problem, we need to find the measure of angle C in triangle ABC, given that AC = BC and AB² = 2AC².

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Triangle ABC is isosceles with AC = BC.
- We have the equation: AB² = 2AC².

2. **Substitute AC for BC:**
- Since AC = BC, we can replace BC in the equation. Thus, we rewrite AB²:
\[ AB² = 2AC² \]

3. **Use the Isosceles Triangle Property:**
- In triangle ABC, since AC = BC, we can denote AC as 'x'. Therefore:
\[ AB² = 2x² \]

4. **Apply the Cosine Rule:**
- In triangle ABC, we can apply the cosine rule:
\[ AB² = AC² + BC² - 2 \cdot AC \cdot BC \cdot \cos(C) \]
Substituting AC and BC with 'x':
\[ AB² = x² + x² - 2x² \cdot \cos(C) \]
Simplifying this gives:
\[ AB² = 2x² - 2x² \cdot \cos(C) \]

5. **Set the Two Expressions for AB² Equal:**
- From the earlier step, we have:
\[ 2x² = 2x² - 2x² \cdot \cos(C) \]
Rearranging this gives:
\[ 0 = -2x² \cdot \cos(C) \]

6. **Solve for cos(C):**
- Since \( -2x² \) cannot be zero (x is a length), we can conclude:
\[ \cos(C) = 0 \]

7. **Determine the Angle C:**
- The angle C for which cos(C) = 0 is:
\[ C = 90^\circ \]

### Conclusion:
The measure of angle C is \( 90^\circ \).