The Bretscheiner’s formula helps find the area of a non-cyclic quadrilateral, that cannot be inscribed in a circle,

using only side lengths and possibly one angle measure or one diagonal length.

After the Bretschneider’s formula, we’ll simplify the quadrilateral to make it cyclic. We’ll get the Brahmagupta’s formula.

With the cyclic quadrilateral the product of the diagonals $e$ and $f$ is $ef=ac+bd$ and opposite angles are supplementary.

Let $K$ be area.

From the figure, we notice $K=A_{ABD}+A_{DCB}$

$K=\frac{1}{2}ad \sin \alpha+ \frac{1}{2}bc \sin \phi$

$2K=ad \sin \alpha+ bc \sin \phi$

Squaring both sides:

$4K^{2}=(ad)^{2} \sin^{2} \alpha+ (bc)^{2} \sin^2 \phi+ 2adbc \sin \alpha \sin \phi$

$4K^{2}=(ad)^{2} \sin^{2} \alpha+ (bc)^{2} \sin^2 \phi+ 2adbc \sin \alpha \sin \phi$ $(1)$

Now let’s check 2 values of diagonal $e$ :

$e^{2}=a^{2}+d^{2}-2ad \cos \alpha$

$e^{2}=b^{2}+c^{2}-2bc \cos \phi$

$a^{2}+d^{2}-2ad \cos \alpha=b^{2}+c^{2}-2bc \cos \phi$

$a^{2}+d^{2}-b^{2}-c^{2}=2ad \cos \alpha-2bc \cos \phi$

$\frac{a^{2}+d^{2}-b^{2}-c^{2}}{2}=ad \cos \alpha-bc \cos \phi$

Taking the square:

$\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad)^{2} \cos^{2} \alpha+(bc)^{2} \cos^{2} \phi-2abcd \cdot \cos \phi \cdot \cos \alpha$ $(2)$

Adding $(1)$ and $(2)$ :

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad)^{2}(\sin^{2} \alpha+\cos^{2} \alpha)+(bc)^{2}(\sin^{2} \phi+\cos^{2} \phi)+ 2adbc \cdot \sin \alpha \sin \phi -2abcd \cdot \cos \alpha \cos \phi$

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad)^{2}+ (bc)^{2}- 2adbc \cdot \cos (\alpha + \phi)$

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad +bc)^{2}-2adbc- 2adbc \cdot \cos (\alpha + \phi)$

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad +bc)^{2}- 2adbc \cdot( \cos (\alpha + \phi)+1)$

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad +bc)^{2}- 4adbc(\frac{\cos (\alpha + \phi)+1}{2})$ Injected 2 in the last term.

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad +bc)^{2}- 4adbc \cdot \cos^{2} (\frac{\alpha + \phi}{2})$

To the same denominator and multiplying both sides by 4:

$16K^{2}=4(ad +bc)^{2}-(a^{2}+d^{2}-b^{2}-c^{2})^{2}- 16adbc \cdot \cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=[4(ad +bc)^{2}]-[(a^{2}+d^{2}-b^{2}-c^{2})^{2}]- 16adbc \cdot \cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=(2ad +2bc+a^{2}+d^{2}-b^{2}-c^{2})(2ad +2bc-a^{2}-d^{2}+b^{2}+c^{2})- 16adbc \cdot\cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=(a^{2}+2ad+d^{2}-(b^{2}-2bc+c^{2}))(-a^{2}+2ad-d^{2}+(b^{2}+2bc+c^{2}))- 16adbc \cdot \cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=((a+d)^{2}-(b-c)^{2})((b+c)^{2}-(a-d)^{2})- 16adbc \cdot\cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=(a+d+b-c)(a+d-b+c)(b+c+a-d)(b+c-a+d)- 16adbc \cdot\cos^{2} (\frac{\alpha + \phi}{2})$

$K^{2}=\frac{b+c+d-a}{2}\cdot \frac{a+c+d-b}{2}\cdot \frac{a+b+d-c}{2}\cdot \frac{a+b+c-d}{2}- adbc\cdot \cos^{2} (\frac{\alpha + \phi}{2})$

$\boxed {K=\sqrt{\frac{b+c+d-a}{2}\cdot \frac{a+c+d-b}{2}\cdot \frac{a+b+d-c}{2}\cdot \frac{a+b+c-d}{2}- adbc\cdot \cos^{2} (\frac{\alpha + \phi}{2})}}$

If we call the semi-perimeter $s=\frac{a+b+c+d}{2}$

We have the final Bretscheiner’s formula:

$\boxed {K=\sqrt{(s-a)(s-b)(s-c)(s-d)-adbc\cdot \cos^{2} (\frac{\alpha + \phi}{2})}}$

Brahmagupta’s formula

The steps remain the same but this time since the quadrilateral is $CYCLIC$ we know that the two opposite angles are supplementary.

$\cos \frac{\pi}{2}=0$

We get from the Bretscheiner’s formula

$\boxed {K=\sqrt{(s-a)(s-b)(s-c)(s-d)}}$

Heron’s formula

The fourth side goes away and we get a triangle. From Brahmagupta’s formula:

$\boxed {K=\sqrt{s(s-a)(s-b)(s-c)}}$

In depth explanation:

From Al-Kashi to Bretshneider via Heron and Brahmagupta, we are going to go through some of their work:

Al-kashi is simply the Law of Cosines:

Law of cosines:

From what we know:
$\cos{A}=\frac{AD}{c} \Rightarrow AD=c\cdot \cos{A}$
We have two right triangles: $\triangle{ADB}$ and $\triangle{CDB}$
Using the Pythagorean Theorem:
${h}^2={c}^2-{AD}^2$
Also:
${h}^2={a}^2-{DC}^2$
${a}^2-{DC}^2={c}^2-{AD}^2$
But :
$\overline{AC}=\overline{AD}+\overline{DC} \Rightarrow \overline{DC}=\overline{AC}-\overline{AD}$
This means:
$DC=AC-AD=b-AD$
${a}^2-{b-AD}^2={c}^2-{AD}^2$
${a}^2-({b}^2-2 \cdot b \cdot AD+{AD}^2) ={c}^2-{AD}^2$
${a}^2-{b}^2+2 \cdot b \cdot AD-{AD}^2={c}^2-{AD}^2$
${a}^2-{b}^2+2 \cdot b \cdot (AD)={c}^2$
We know from above: $AD=c\cdot\cos{A}$
${a}^2-{b}^2+2 \cdot b \cdot c\cdot\cos{A}={c}^2$
Finally we have the $Law \;of\; cosines$ :

${a}^2={b}^2+{c}^2-2 \cdot b \cdot c\cdot\cos{A}$

${b}^2={a}^2+{c}^2-2 \cdot a \cdot c\cdot\cos{B}$

${c}^2={a}^2+{b}^2-2 \cdot a \cdot b\cdot\cos{C}$

Heron Formula:

This one is widely used. We use it when we have only sides and we want to find the Area of the triangle

Let $S$ be area and $s=\frac{a+b+c}{2}$ the semi-perimeter.

$S=\frac{1}{2}bc \sin A$ .

We should remember this because we’ll need it in the following paragraphs.

$\sin A=\frac{2S}{bc}$

$a^2=b^2+c^2-2bc \cos A \Rightarrow \cos A=\frac{b^2+c^2-a^2}{2bc}$

But we know that:

$\sin^{2} A+ \cos^{2} A=1$

$(\frac{2S}{bc})^{2}+(\frac{b^2+c^2-a^2}{2bc})^{2}=1$

$\frac{4S^{2}}{b^{2}c^{2}}+\frac{(b^2+c^2-a^2)^{2}}{4b^{2}c^{2}}=1$

$\frac{4\times 4S^{2}}{4 \times b^{2}c^{2}}+\frac{(b^2+c^2-a^2)^{2}}{4b^{2}c^{2}}=\frac{4b^{2}c^{2}}{4b^{2}c^{2}}$

$16S^{2}=4b^{2}c^{2}-(b^2+c^2-a^2)^{2}$

This is a difference of squares, we can factor:

$16S^{2}=(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)$

$16S^{2}=[(b+c)^2-a^2][a^2-(b-c)^{2}]$

$16S^{2}=(b+c-a)(b+c+a)(a+b-c)(a+c-b)$

$16S^{2}=(a+b+c)(b+c-a)(a+c-b)(a+b-c)$

$S^{2}=\frac{(a+b+c)}{2}\frac{(b+c-a)}{2}\frac{(a+c-b)}{2}\frac{(a+b-c)}{2}$

$S=\sqrt{s(s-a)(s-b)(s-c)}$

This is the Heron formula:

$\boxed{S=\sqrt{s(s-a)(s-b)(s-c)}}$

The Bretschneider formula:

We will demonstrate the following forms of the Area $K$ per BRETSCHNEIDER

$K=\frac{1}{2}pq \sin \theta$

$K=\frac{1}{4}(b^{2}+d^{2}-a^{2}-c^{2}) \tan \theta$

$K=\frac{1}{4}\sqrt{4p^{2}q^{2}-(b^{2}+d^{2}-a^{2}-b^{2})^{2}}$

We split the diagonal $p$ into $(p-y)$ and $y$

The same way for $q$ : $(q-x)$ and $x$

We express the area of the four triangles:

$A_1=\frac{1}{2}xy \sin \theta$

$A_2=\frac{1}{2}x(p-y) \sin \theta$

$A_3=\frac{1}{2}(p-y)(q-x) \sin \theta$

$A_4=\frac{1}{2}(q-x)y \sin \theta$

Now we add all four areas:

$A_{ABCD}=A_1+A_2+A_3+A_4$

$A_{ABCD}=(\frac{1}{2}xy+\frac{1}{2}px-\frac{1}{2}xy+\frac{1}{2}pq-\frac{1}{2}px-\frac{1}{2}qy+\frac{1}{2}xy+\frac{1}{2}qy-\frac{1}{2}xy) \sin \theta$

After simplication:

$A_{ABCD}=\frac{1}{2}pq \sin \theta$

The first form:

$\boxed{K=\frac{1}{2}pq \sin \theta}$

Now the second form:

From the figure:

$c^{2}=x^{2}+y^{2}-2xy \cos \theta$ $(1)$

$a^{2}=(p-y)^{2}+(q-x)^{2}-2(p-y)(q-x)\cos \theta$

$a^{2}=p^{2}-2py+y^{2}+q^{2}-2qx+x^{2}-(pq-px-qy+xy)\cdot 2\cos \theta$

We expand and replace terms in $(1)$ by $c^{2}$

$a^{2}=p^{2}+c^{2}+q^{2}-2py-2qx-2pq\cos \theta+2px \cos \theta+ 2qy \cos \theta$

Now we move to the $b^{2}$

$b^{2}=(p-y)^{2}+x^{2}+2(p-y)x \cos \theta$

$b^{2}=p^{2}-2py+y^{2}+x^{2}+2px \cos \theta-2xy \cos \theta$

We expand and replace terms in $(1)$ by $c^{2}$

$b^{2}=p^{2}-2py+c^{2}+2px \cos \theta$

Now we move to the $d^{2}$

$d^{2}=(q-x)^{2}+y^{2}+2(q-x)y \cos \theta$

$d^{2}=q^{2}-2qx+x^{2}+y^{2}+2qy\cos \theta-2xy \cos \theta$

$d^{2}=q^{2}-2qx+c^{2}+2qy \cos \theta$

Let’s add as follows:

$b^{2}+d^{2}-a^{2}-c^{2}=$

$=p^{2}-2py+c^{2}+2px \cos \theta+q^{2}-2qx+c^{2}+2qy \cos \theta-p^{2}-c^{2}-q^{2}+2py+2qx+2pq\cos \theta$

$-2px \cos \theta- 2qy \cos \theta-c^{2}$

After simplification we get:

$b^{2}+d^{2}-a^{2}-c^{2}=2pq \cos \theta \Rightarrow pq=\frac{b^{2}+d^{2}-a^{2}-c^{2}}{2\cos \theta}$

But:

$K=\frac{1}{2}pq \sin \theta$

We plug in $pq$

$K=\frac{1}{4}(b^{2}+d^{2}-a^{2}-b^{2}) \tan \theta$

Second Form:

$\boxed{K=\frac{1}{4}(b^{2}+d^{2}-a^{2}-c^{2}) \tan \theta}$

Third form:

From the sequences above:

$K=\frac{1}{2}pq \sin \theta$

But: $\sin \theta=\sqrt{1- \cos^{2} \theta}$

$K=\frac{1}{2}pq\sqrt{1-(\frac{b^{2}+d^{2}-a^{2}-c^{2}}{2pq})^{2}}$

$K=\frac{1}{2}pq\sqrt{1-\frac{(b^{2}+d^{2}-a^{2}-c^{2})^{2}}{4p^{2}q^{2}}}$

$K=\frac{1}{2}\sqrt{p^{2}q^{2}-p^{2}q^{2}\frac{(b^{2}+d^{2}-a^{2}-c^{2})^{2}}{4p^{2}q^{2}}}$

$K=\frac{1}{4}\sqrt{4p^{2}q^{2}-(b^{2}+d^{2}-a^{2}-c^{2})^{2}}$

This is the actual BRETSCHNEIDER formula:

$\boxed{K=\frac{1}{4}\sqrt{4p^{2}q^{2}-(b^{2}+d^{2}-a^{2}-c^{2})^{2}}}$

According to wiki, the following work is credited to COOLIDGE in 1939:

Let $K$ be area.

From the figure, we notice $K=A_{ABD}+A_{DCB}$

$K=\frac{1}{2}ad \sin \alpha+ \frac{1}{2}bc \sin \phi$

$2K=ad \sin \alpha+ bc \sin \phi$

Squaring both sides:

$4K^{2}=(ad)^{2} \sin^{2} \alpha+ (bc)^{2} \sin^2 \phi+ 2adbc \sin \alpha \sin \phi$

$4K^{2}=(ad)^{2} \sin^{2} \alpha+ (bc)^{2} \sin^2 \phi+ 2adbc \sin \alpha \sin \phi$ $(1)$

Now let’s check 2 values of diagonal $e$ :

$e^{2}=a^{2}+d^{2}-2ad \cos \alpha$

$e^{2}=b^{2}+c^{2}-2bc \cos \phi$

$a^{2}+d^{2}-2ad \cos \alpha=b^{2}+c^{2}-2bc \cos \phi$

$a^{2}+d^{2}-b^{2}-c^{2}=2ad \cos \alpha-2bc \cos \phi$

$\frac{a^{2}+d^{2}-b^{2}-c^{2}}{2}=ad \cos \alpha-bc \cos \phi$

Taking the square:

$\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad)^{2} \cos^{2} \alpha+(bc)^{2} \cos^{2} \phi-2abcd \cdot \cos \phi \cdot \cos \alpha$ $(2)$

Adding $(1)$ and $(2)$ :

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=$

$=(ad)^{2}(\sin^{2} \alpha+\cos^{2} \alpha)+(bc)^{2}(\sin^{2} \phi+\cos^{2} \phi)+ 2adbc \cdot \sin \alpha \sin \phi -2abcd \cdot \cos \alpha \cos \phi$