<s>[INST] Maximum value of diagonals angle in parallelogram Let $ABCD$ be a parallelogram with sides a and b $(a>b)$ and let $r=a/b$ the ratio of the two sides.
The two diagonals form 4 angles which are by two congruent. Let's call the acute angle φ and the obtuse θ.
Find the maximum value that φ can take, in relation to r.
I have managed to find a relationship for the angles φ and θ, but they also contain the diagonals:
$\cosφ = (D_1^2+D_2^2-4b^2)/2D_1D_2$ and $\cos θ = (D_1^2+D_2^2-4a^2)/2D_1D_2$ but I don't know how to get rid of the diagonals and somehow take into account the ratio a/b. 
 [/INST] Let $\alpha$ be a measure of the angle of our parallelogram. 
Thus, $S_{ABCD}=ab\sin\alpha=rb^2\sin\alpha$.
In another hand, by law of cosines we obtain:
$$a^2=\frac{AC^2}{4}+\frac{BC^2}{4}-\frac{AC\cdot BC\cos(180^{\circ}-\varphi)}{2}$$ and 
$$b^2=\frac{AC^2}{4}+\frac{BC^2}{4}-\frac{AC\cdot BC\cos\varphi}{2},$$ 
which gives
$$a^2-b^2=AC\cdot BC\cos\varphi$$ or
$$AC\cdot BC=\frac{a^2-b^2}{\cos\varphi}$$ and
$$S_{ABCD}=\frac{(a^2-b^2)\tan\varphi}{2}=\frac{b^2(r^2-1)\tan\varphi}{2}.$$
Id est, $$\frac{b^2(r^2-1)\tan\varphi}{2}=rb^2\sin\alpha$$
or $$\tan\varphi=\frac{2r\sin\alpha}{r^2-1}.$$
We see that $0^{\circ}<\varphi<90^{\circ}$ and $\phi$ can get all these values,
which says that the maximum does not exist and the supremum is $90^{\circ}$.
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