Let ABC be a triangle such that angle ACB = 135 degrees. Prove that
AB^2 = AC^2 + BC^2 + ¡Ì2 x AC x BC.
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To prove the given statement, we will make use of the Law of Cosines, which relates the lengths of the sides of a triangle to the cosine of one of its angles.
The Law of Cosines states that in a triangle with sides a, b, and c, and the angle between sides a and b denoted as angle C, the following equation holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
In the given triangle ABC, we need to prove that:
AB^2 = AC^2 + BC^2 + √2 * AC * BC
Let's start the proof:
1. Begin with the Law of Cosines equation for side AB, which is the longest side of the triangle.
AB^2 = AC^2 + BC^2 - 2AC * BC * cos(angle ACB)
2. Since it is given that angle ACB = 135 degrees, substitute the value into the equation:
AB^2 = AC^2 + BC^2 - 2AC * BC * cos(135°)
3. We know that cos(135°) = -sqrt(2)/2 according to the unit circle or reference angles.
Substitute this value into the equation:
AB^2 = AC^2 + BC^2 - 2AC * BC * (-sqrt(2)/2)
4. Simplify the equation and distribute the negative sign:
AB^2 = AC^2 + BC^2 + sqrt(2) * AC * BC
5. Thus, the equation is proven, and we have shown that AB^2 = AC^2 + BC^2 + sqrt(2) * AC * BC.
This completes the proof.