1. **Stating the problem:** We are given six expressions involving variables $d$ and $b$: $$a_1 = \frac{d^2}{4}, \quad a_2 = \frac{d}{2}b, \quad a_3 = \frac{b^2}{4}, \quad a_4 = b, \quad a_5 = d, \quad a_6 = 1$$ There is a right triangle with angle $\alpha$, opposite side $b$, adjacent side $d$, and hypotenuse $1$. 2. **Using the Pythagorean theorem:** Since the triangle is right-angled, the sides satisfy: $$d^2 + b^2 = 1^2 = 1$$ 3. **Relating the expressions:** Notice that $a_1 + a_3 = \frac{d^2}{4} + \frac{b^2}{4} = \frac{d^2 + b^2}{4} = \frac{1}{4}$. 4. **Interpreting the expressions:** - $a_1$, $a_2$, and $a_3$ resemble parts of the expansion of $(\frac{d}{2} + \frac{b}{2})^2$: $$\left(\frac{d}{2} + \frac{b}{2}\right)^2 = \frac{d^2}{4} + \frac{d}{2}b + \frac{b^2}{4} = a_1 + a_2 + a_3$$ 5. **Summary:** - The sum $a_1 + a_2 + a_3$ equals $\left(\frac{d}{2} + \frac{b}{2}\right)^2$. - The Pythagorean relation $d^2 + b^2 = 1$ holds. This confirms the geometric relations in the triangle and the algebraic expressions given. **Final key relation:** $$a_1 + a_3 = \frac{1}{4}$$