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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?

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Did you check out the references (4 and 11) in the wiki page here: en.wikipedia.org/wiki/Cyclic_quadrilateral? –  Aryabhata Jan 26 '12 at 19:51
There are also some relevant formulas with some derivations at mathworld.wolfram.com/CyclicQuadrilateral.html –  Isaac Feb 3 '12 at 4:12
If you like, you can prove the first theorem via brute force with trig. Assuming the circumdiameter is $k$, the Law of Sines allows you to write $$a = k\sin\angle BCA \qquad b = k\sin\angle BAC \qquad c = k\sin\angle DAC \qquad d = k\sin\angle DCA$$ Also, $$|AC| = k\sin B = k\sin D = k\sin(\alpha+\beta) = k\sin(\gamma+\delta)$$ Opposite angles being supplementary, we have $$\delta = \pi - \alpha - \beta - \gamma$$ Substitute into the ostensible formula $$|AC|^2 ( a b + c d ) = ( a c + b d )( a d + b c )$$, expand, and simplify like mad to verify the equality. –  Blue Oct 27 '13 at 14:51
Of course, the same brute force trig technique works on Theorem 2, as well. –  Blue Oct 27 '13 at 15:02

1 Answer 1

up vote 3 down vote accepted

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)


Thus $$R=\frac{(ab+cd)AC}{4\sqrt{(p-a)(p-b)(p-c)(p-d)}}=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(p-a)(p-b)(p-c)(p-d)}}$$

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