Problem : Let $ABCD$ be a convex quadrilateral sides : $a,b,c,d$ and $e,f$ sides of diagonal : Prove that :
$$16S^{2}+\left(a^{2}+c^{2}-b^{2}-d^{2}\right)^{2}=4e^{2}f^{2}$$
Where : $[ABCD]=S$ area
I don't know how I starte in this problem ? I'm thinking use $\operatorname{altitude}$
Label the intersection of the diagonal $O$, and let $AO = p, BO = q, CO = r, DO = s, $ and angle $AOB = DOC = \theta. \ $ Note that angle $BOC = $ angle $DOA = 180 - \theta.$ I'm also labelling sides $AB$ as $a$, $BC \ b$, $CD \ c$, and $DA \ a$, for readability.
Using the cosine rule on triangles $AOB, \ BOC, \ DOC \ and \ DOA$ respectively, we get: $$a^2 = p^2 + q^2 -2pq \ cos\theta \quad (1) \\ b^2 = q^2 + r^2 -2qr \ cos (180-\theta) \quad (2) \\ c^2 = r^2 + s^2 -2rs \ cos\theta \quad (3) \\ d^2 = p^2 + s^2 -2ps \ cos (180-\theta) \quad (4) \\ $$
Using the fact that $cos (180-\theta) = - cos\theta, \ $ (1) + (3) - (2) - (4) then gives:
$a^2 + c^2 - b^2 - d^2 = -2(p+r)(q+s) \ cos\theta = -2ef cos\theta, \ $ and so
$(a^2 + c^2 - b^2 - d^2)^2 = 4e^2f^2 \ cos^2\theta \quad (5)$
Now, we can also write the area of ABCD using the Area Sine Law for each of the 4 triangles, and using the fact that $sin(180-\theta) = sin\theta , \ $ and after some further manipulation similar to above, we get:
$S = Area ABCD = \frac{1}{2}ef sin\theta, \ $ therefore:
$16S^2 = 4e^2f^2sin^2\theta \quad (6).$
$(5) + (6)$ gets the result.