: 400px Let the area of $\triangle DAB$ and $\triangle BCD$ be $\mathcal A_1$ and $\mathcal A_2$. From Area of Triangle in Terms of Two Sides and Angle: :$\mathcal A_1 = \dfrac {a b \sin \alpha} 2$ and $\mathcal A_2 = \dfrac {c d \sin \gamma} 2$ From to the second axiom of area, $\mathcal A = \mathcal A_1 + \mathcal A_2$, so: {{begin-eqn}} {{eqn | l = \mathcal A^2       | r = \frac 1 4 \left({a^2 b^2 \sin^2 \alpha + 2 a b c d \sin \alpha \sin \gamma + c^2 d^2 \sin^2 \gamma}\right)       | n = 1 }} {{end-eqn}} The diagonal $p$ can be written in 2 ways using the Law of Cosines: :$p^2 = a^2 + b^2 - 2 a b \cos \alpha$ :$p^2 = c^2 + d^2 - 2 c d \cos \gamma$ Equality is transitive, so: {{begin-eqn}} {{eqn | l = a^2 + b^2 - 2 a b \cos \alpha       | r = c^2 + d^2 - 2 c d \cos \gamma }} {{eqn | l = a^2 + b^2 - c^2 - d^2       | r = 2 a b \cos \alpha - 2 c d \cos \gamma       | c = adding $2 a b \cos \alpha - c^2 - d^2$ to both sides }} {{eqn | l = \left({a^2 + b^2 - c^2 - d^2}\right)^2       | r = 4 a^2 b^2 \cos^2 \alpha - 8 a b c d \cos \alpha \cos \gamma + 4 c^2 d^2 \cos^2 \gamma       | c = squaring both sides }} {{eqn | l = 0       | r = \frac 1 4 \left({a^2 b^2 \cos^2 \alpha - 2 a b c d \cos \alpha \cos \gamma + c^2 d^2 \cos^2 \gamma}\right)       | c = algebraic manipulation }} {{eqn | ro= -       | r = \frac 1 {16} \left({a^2 + b^2 - c^2 - d^2}\right)^2       | o =  }} {{end-eqn}} Now add this equation to $(1)$. Then trigonometric identities can be used, as follows: {{begin-eqn}} {{eqn | l = \mathcal A^2       | r = \frac 1 4 \left({a^2 b^2 + c^2 d^2 - 2 a b c d \cos \left({\alpha + \gamma}\right)}\right) - \frac 1 {16} \left({a^2 + b^2 - c^2 - d^2}\right)^2       | c = Sum of Squares of Sine and Cosine and Cosine of Sum }} {{eqn | r = \frac 1 {16} \left({4 a^2 b^2 + 4 c^2 d^2 - \left({a^2 + b^2 - c^2 - d^2}\right)^2}\right) - \frac 1 2 a b c d \cdot \cos \left({\alpha + \gamma}\right) }} {{end-eqn}} By expanding the square $\left({a^2 + b^2 - c^2 - d^2}\right)^2$:  {{begin-eqn}} {{eqn | n = 2       | l = \mathcal A^2       | r = \frac 1 {16} \left({-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2}\right) - \frac 1 2 abcd \cos \left({\alpha + \gamma}\right) }} {{end-eqn}} Adding and subtracting $8 a b c d$ to and from the numerator of the first term of $(2)$: {{begin-eqn}} {{eqn | l = \mathcal A^2       | r = \frac 1 {16} \left({-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2 + 8 a b c d - 8 a b c d}\right) - \frac 1 2 a b c d \cos \left({\alpha + \gamma}\right) }} {{end-eqn}} allows the product $\left({-a + b + c + d}\right) \left({a - b + c + d}\right) \left({a + b - c + d}\right) \left({a + b + c - d}\right)$ to be formed: {{begin-eqn}} {{eqn | l = \mathcal A^2       | r = \frac 1 {16} \left({-a + b + c + d}\right) \left({a - b + c + d}\right) \left({a + b - c + d}\right) \left({a + b + c - d}\right)       | c =  }} {{eqn | ro= -       | r = \frac 1 2 a b c d - \frac 1 2 a b c d \cos \left({\alpha + \gamma}\right)       | o =  }} {{eqn | r = \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - \frac 1 2 a b c d - \frac 1 2 a b c d \cos \left({\alpha + \gamma}\right)       | c = as $s = \dfrac {a + b + c + d} 2$ }} {{eqn | r = \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - \frac 1 2 a b c d \left({1 + \cos \left({\alpha + \gamma}\right)}\right)       | c =  }} {{eqn | r = \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - a b c d \cos^2 \left({\dfrac {\alpha + \gamma} 2} \right)       | c = Half Angle Formula for Cosine }} {{end-eqn}} Hence the result. {{qed}}