I want to know ifthink this methodis a potentially new formula for finding the area of a general quadrilateral with. Also do note this post has been repurposed for combining all my 4 posts regarding my formula as they were being closed for being duplicates. This post might have some linguistic errors or repeated information, please let me know if there is and I'll edit them
Case 1: A convex quadrilateral
In the sidesabove convex non intersecting quadrilateral we take AB=a, BC=b, CD=c, AD=d and an angle being given$\angle ABC$ is correct or not. 
$\theta$.
In$$ Area of ABCD \\ = Ar\triangle ABC+Ar\triangle ACD \\ =\frac{1}{2}ab\sin\theta+\frac{1}{2}cd\sin\angle D \\ $$ Now the problem is to find $ \triangle ABC $$\sin\angle D$, using law of cosines we haveget: $$ AC^2 = AB^2 + BC^2 -2(AB)(BC)\cos\angle ABC $$$$ AC^2=a^2+b^2-2ab\cos\theta \\ and AC^2=c^2+d^2-2cd\cos\angle D $$ Subtracting both equations from each other we get: $$ AC^2-AC^2=(a^2+b^2-2ab\cos\theta)-(c^2+d^2-2cd\cos\angle D), \\ a^2+b^2-c^2-d^2-2ab\cos\theta+2cd\cos\angle D=0, \\ \cos\angle D= \frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd} $$ Now thatusing the trigonometric identity $ \sin\angle D= \sqrt{1-\cos^2\angle D} $ we knowget: $$ \sin\angle D= \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} $$ And putting this into $ \frac{1}{2}ab\sin\theta+\frac{1}{2}cd\sin\angle D $ we get: $$ \frac{1}{2}\left[ab \sin \theta + cd \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} \right] $$ Which I believe to be the valuesarea of ABany non intersecting convex quadrilateral. Please correct me if I'm wrong at any step. Also, BCdo note that the a and b are the sides which are adjacent to the given angle, CDmeaning that the lines a and b are the ones which make the given angle
Case 2: A concave quadrilateral
We will be taking $AB=a$, AD$BC=b$, $CD=c$, $AD=d$ and AC$reflex\angle ABC$ as $\theta$
$$
Ar\triangle ADC=\frac{1}{2}cd\sin \angle D \\ Ar\triangle ABC=\frac{1}{2}ab\sin (360-\theta)=\frac{-1}{2}ab\sin \theta\\
$$
Now using the law of cosines:
$$
AC^2=a^2+b^2-2ab\cos (360-\theta) \\ which is AC^2=a^2+b^2-2ab\cos \theta \\
AC^2=c^2+d^2-2cd\cos \angle D
$$
Subtracting both equations from each other:
$$
AC^2-AC^2=(a^2+b^2-2ab\cos\theta)-(c^2+d^2-2cd\cos\angle D), \\ a^2+b^2-c^2-d^2-2ab\cos\theta+2cd\cos\angle D=0, \\
\cos\angle D= \frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}
$$
Now using the trigonometric identity $ \sin\angle D= \sqrt{1-\cos^2\angle D} $ we get:
$$
\sin\angle D= \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2}
$$
now here comes the final step:
$$
ArABCD=Ar\triangle ADC-Ar\triangle ABC \\
=\frac{1}{2}cd\sin \angle D-\frac{-1}{2}ab\sin \theta \\
=\frac{1}{2}\left[cd \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2}+ab\sin \theta \right]
=\frac{1}{2}\left[ab \sin \theta + cd \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} \right]
$$
This is the proof that my formula works for any concave quadrilateral when the reflex angle is given. A similar proof can findbe done in case the areasgiven angle is opposite to the reflex angle. If the given angle is neither the reflex angle nor the one opposite to the reflex angle the proof is similar to the proof of the formula for any convex quadrilateral, whose post I have linked. Please correct me if I'm wrong in any step.
Case 3: A Self Intersecting Quadrilateral
Here ABCD is our self intersecting quadrilateral whose area will be denoted by $ \triangle ABC $$ \Delta $. We will be taking $AB=a, BC=c, CD=c, AD=d$ and $ \triangle ACD $ using Heron's formula$\angle B$ as $\theta$. Adding their areasWe are assuming that $a,b,c,d,\theta$ are the given sides and angle. Also we gethave to be careful while assigning the sides these variables, as the angle $\theta$ must lie between the sides $a,b$ $$ \Delta = Ar\triangle ADC+Ar\triangle ABC \\ \Delta =\frac{1}{2}cd\sin \angle D+\frac{1}{2}ab\sin \theta $$ Do note that the intersecting area of quadrilateral ABCD$\triangle AEC$ is accounted by $\sin \theta$ as it can be negative when $\theta$ is reflex. Now by using the law of cosine: My question here is if$$ AC^2=a^2+b^2-2ab\cos \theta \\ AC^2=c^2+d^2-2cd\cos \angle D $$ Subtracting both of them from each other we get: $$ a^2+b^2-2ab\cos \theta-c^2-d^2+2cd\cos \angle D=0 \\ \cos \angle D=\frac{c^2+d^2-a^2-b^2+2ab\cos \theta}{2cd} $$ Now using the trigonometric identity $ \sin\angle D= \sqrt{1-\cos^2\angle D} $ we get: $$ \sin\angle D= \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} $$ And putting this into $ \frac{1}{2}ab\sin\theta+\frac{1}{2}cd\sin\angle D $ we get: $$ \frac{1}{2}\left[ab \sin \theta + cd \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} \right] $$ So my approach wasformula is correct for any self intersecting quadrilateral,and as seen in the linked posts,it is also correct for any concave and convex quadrilateral. Essentially making it a formula for a general quadrilateral which requires only 4 sides and 1 angle. Please let me know if yesI'm wrong.
The more important thing here is that the formula may be very tedious and more time consuming than the method itself, whether orbut the fact that it exists shows that the area of a quadrilateral will always remain same with the same 4 sides and one angle, as long as the given angle lies between two sides of the same length. Which I think was believed to not be the case previously.
Further this formula is a known resultmore minimalistic than Bretschneider's formula, as it requires one angle instead of two.
