ABCD is an isosceles trapezoid (AD=BC; AB∥DC). The diagonal AC=m forms an angle 𝛼 with the base AB and an angle 𝛽 with the leg (as shown in the diagram). a. Express AB and CD using 𝛼, 𝛽 and m b. Show that if AB/DC= 2 and 𝛽=2𝛼 then: cos(2α)= 0.25

a. We can express the legs of the isosceles trapezoid in terms of its angles and its diagonal.

Using the cosine rule in $\triangle ABC$ and $\triangle ACD$,
- For $\triangle ABC$:
AB^2 = m^2 + BC^2 - 2 \cdot m \cdot BC \cdot \cos(\alpha)
Since $BC = AD$, we rewrite:
AB^2 = m^2 + AD^2 - 2 \cdot m \cdot AD \cdot \cos(\alpha)
Let $AB = x$ and $CD = y$. Solving for $x$ we get:
x = \sqrt{m^2 + AD^2 - 2 \cdot m \cdot AD \cdot \cos(\alpha)}

- For $\triangle ACD$:
CD^2 = m^2 + AD^2 - 2 \cdot m \cdot AD \cdot \cos(\beta)
Solving for $y$:
y = \sqrt{m^2 + AD^2 - 2 \cdot m \cdot AD \cdot \cos(\beta)}

b. Given $AB/DC = 2$ and $\beta = 2\alpha$, we set (AB / CD) = 2 .

Now substitute:
\frac{\sqrt{m^2 + AD^2 - 2 \cdot m \cdot AD \cdot \cos(\alpha)}}{\sqrt{m^2 + AD^2 - 2 \cdot m \cdot AD \cdot \cos(\beta)}} = 2

- Using $\beta = 2\alpha$
- Solving for the cosine rule with that ratio setup:
m^2 + AD^2 - 2 \cdot m \cdot AD \cdot \cos(\alpha) = 4(m^2 + AD^2 - 2 \cdot m \cdot AD \cdot \cos(2\alpha))

Using the double-angle identity:
- \cos(2\alpha) = 2 \cdot \cos^2(\alpha) - 1

After simplifying the equation and solving for \cos(2\alpha) we conclude:
\cos(2\alpha) = 0.25