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If $ABCD$ is a cyclic quadrilateral, then

$$

AC\cdot(AB\cdot BC+CD\cdot DA)=BD\cdot (DA\cdot AB+BC\cdot CD)

$$

I tried using many approaches, but I could not find a proper solution. Can anyone please help me with this problem?

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Let the radius of the circle around the quadrilateral be $R$. Consider the diagonal $AC$. Drop perpendicular on $AC$ from $B$. Let, the point be $M$, where there perpendicular meets $AC$. Consider another perpendicular from $D$ on $AC$. Let, the perpendicular meet $AC$ at $N$. Now,

$AB.BC=2R.BM$ [Brahmagupta](I hope you know that,do let me know if you don’t know it)

$DA.DC=2R.DN$ [Brahmagupta]

Now,

$AC.[AB.BC+CD.DA]=AC.2R.BM+AC.2R.DN$

So,$AC.[AB.BC+CD.DA]=2R[2\Delta ABC+2\Delta ACD]$

Thus, $AC.[AB.BC+CD.DA]=4RS$[S=Area of ABCD]

Similarly, you can show that,

$BD.[DA.AB+BC.CD]=4RS$

Hence, Proved.

Note that, as a consequence of the Law of Sines, a triangle $\triangle ABC$ with circumdiameter $d$ has sides of length

$$|\overline{BC}| = d\sin A \qquad |\overline{CA}| = d\sin B \qquad |\overline{AB}| = d\sin C$$

The product of a triangle’s three side-lengths, then, is nicely proportional to its area

$$|\overline{AB}||\overline{BC}||\overline{CA}| = 2 d \cdot \frac{1}{2}|\overline{AB}||\overline{BC}|\sin B = 2 d \; |\triangle ABC| \qquad (\star)$$

With $d$ the circumdiameter of the cyclic quadrilateral $\square ABCD$, the relation to be proven easily reduces via $(\star)$ to

$$2d \; \left(\; |\triangle ABC| + |\triangle ADC| \;\right) = 2 d \; \left(\; |\triangle BAD| + |\triangle BCD| \;\right)$$

and we see that each sum of triangle areas simply gives the area of $\square ABCD$ itself.

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