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.

• Maximum angle is 90°, because a rectangle with sides $a$ and $b$ always exists. – Aretino Oct 19 '17 at 17:04

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 $90^{\circ}$.
• @Sal.Cognato All parallelogram has four angles with equal sinuses. If one of measures angles is $\alpha$ then the rest they are $\alpha$, $180^{\circ}-\alpha$ and $180^{\circ}-\alpha$. – Michael Rozenberg Oct 19 '17 at 22:38
Here is a picture of two parallelograms with equal side lengths $a,b$
At the limit how small can $\phi$ be?