In the diagram below, D = α, C1 = β, AB = y, AD = CD = x and AB ⊥ BC. Prove that y = xsinβ√2(1-cosa).

We are given that $AD = CD = x$ and $\hat{D} = \alpha$. Since $AD = CD$, $\triangle ADC$ is an isosceles triangle.
Using the Cosine Rule to find the length of side $AC$:

$$AC^2 = AD^2 + CD^2 - 2(AD)(CD) \cos \hat{D}$$

Substitute the given values:

$$AC^2 = x^2 + x^2 - 2(x)(x) \cos \alpha$$ $$AC^2 = 2x^2 - 2x^2 \cos \alpha$$

Factor out $2x^2$:

$$AC^2 = 2x^2 (1 - \cos \alpha)$$

Take the square root of both sides to find $AC$:

$$AC = \sqrt{2x^2 (1 - \cos \alpha)}$$ $$AC = x \sqrt{2(1 - \cos \alpha)}$$

We are given that $AB \perp BC$, so $\triangle ABC$ is a right-angled triangle at B.
We have $AB = y$ and $\hat{C}_1 = \beta$, which is $\angle ACB$.
In a right-angled triangle, the sine of an angle is the ratio of the opposite side to the hypotenuse:

$$\sin \hat{C}_1 = \frac{AB}{AC}$$

Substitute the given values:

$$\sin \beta = \frac{y}{AC}$$

Rearrange to solve for $y$:

$$y = AC \sin \beta$$

Substitute $AC = x \sqrt{2(1 - \cos \alpha)}$ from Step 1 into $y = AC \sin \beta$:

$$y = \left(x \sqrt{2(1 - \cos \alpha)}\right) \sin \beta$$ $$y = x \sin \beta \sqrt{2(1 - \cos \alpha)}$$

This proves the statement.