Two theorems about an inscribed quadrilateral and the radius of the circle containing its vertices I think those two theorem are two of the most complicated formulas I have ever seen; please prove it because I am not able to find proofs on the internet:
It is known that if the sides of an inscribed quadrilateral $ABCD$ (that is in the order $AB,BC,CD,DA$) have lengths $a,b,c,d$ respectively and $p$ is the semi perimeter of the quadrilatral, then:

Theorem 1: The length of diagonal $AC$ of the quadrilatral is equal to $$\sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}}\;.$$
Theorem 2: The radius of the circle that contains all the vertices of the quadrilateral is equal to $$\frac14\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(p-a)(p-b)(p-c)(p-d)}}\;.$$

By the way, has anyone seen those theorems in a geometry textbook with solution?

Using cosine rule on triangles $ABC$ and $ACD$ gives $$\frac{a^2+b^2-AC^2}{2ab}+\frac{c^2+d^2-AC^2}{2cd}=\cos(\angle{ABC})+\cos(\angle{CDA})=0$$
Thus $(ab+cd)AC^2=(a^2+b^2)cd+(c^2+d^2)ab=(ac+bd)(ad+bc)$ so $AC=\sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}}$.
The area of a triangle with sides $a, b, c$ and circumradius $R$ is $\frac{abc}{4R}$, so we have by Brahmagupta's formula (where $A$ is the area of the cyclic quadrilateral)
$$\sqrt{(p-a)(p-b)(p-c)(p-d)}=A=\frac{ab(AC)}{4R}+\frac{cd(AC)}{4R}=\frac{(ab+cd)AC}{4}\frac{1}{R}$$
Thus $$R=\frac{(ab+cd)AC}{4\sqrt{(p-a)(p-b)(p-c)(p-d)}}=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(p-a)(p-b)(p-c)(p-d)}}$$
