Lemma Let $ABCD$ be a cyclic quadrilateral. The following are equivalent. (a) $AB \cdot CD = BC \cdot DA$. (b) $AC$ is an $A$-symmedian of $\triangle DAB$. (c) $AC$ is a $C$-symmedian of $\triangle BCD$. (d) $BD$ is a $B$-symmedian of $\triangle ABC$ (e) $BD$ is a $D$-symmedian of $\triangle CDA$. Refer to Euclidean Geometry in Mathematical Olympiads by Evan Chem (Lemma 4.27) By Tanegnt Secant Theorem it suffice to show that $\angle B'AC= \angle ADB'$ But since $B'$ is the reflection of $B$ over $AC$, $\angle BA'C=\angle BAC$. Now Observe that in cyclic quadrilateral $ABCD$, $\angle BAC=\angle BDC$, hence it finally suffice to show that $\angle BDC=\angle ADB'$, which is equivalent to proving $DB'$ and $DB$ are isogonal in $\triangle ADB$ Clearly using the lemma $DB$ is the symmedian in $\triangle ADC$, so we have to show that $DB'$ is the median through $D$ in $\triangle ADC$ which equivalent to proving $G$ midpoint of $AC$ Let angle bisector of $\angle ADC$ intersect $AC$ at $Q$. By angle bisector theorem we know that $\dfrac{AD}{DC}=\dfrac{AQ}{QC}$. But from the problem, we have $\dfrac{AD}{DC}=\dfrac{AB}{BC}=\dfrac{AB'}{B'C}$ $\implies D,B',Q,B$ all lie of $D-$ Applonius circle of $\triangle ADC \implies D,Q,B,B'$ are concyclic. Clearly since $B$ and $B'$ are reflections across $AC$ we have $BQ=B'Q$ and $\overset{\huge\frown}{BQ}=\overset{\huge\frown}{B'Q} \implies DQ$ is angle bisector of $\angle BDB' \implies \angle B'DQ=\angle BDQ$ The fact that $\angle B'DQ=\angle BDQ$ and $\angle ADQ=\angle CDQ$ together give us that $\angle ADB'=\angle CDB$ $\implies B'D $ and $BD$ are isogonal in $\triangle ADC \iff B'D$ is median through $D$ in $\triangle ADC$ (Since $DB$ is symmedian) $\implies G$ is the midpoint of $AC$

Sharygin P18 wrote: The products of the opposite sidelengths of a cyclic quadrilateral $ABCD$ are equal. Let $B'$ be the reflection of $B$ about $AC$. Prove that the circle passing through $A,B', D$ touches $AC$ Since $$ AB\times CD=BC\times AD\implies \frac{AB}{BC}\times \frac{CD}{AD}=1\implies (A,C;B,D)=-1.$$So $ABCD$ is a harmonic quadrilateral. Lemma: Let $ABDC$ be an harmonic quadrilateral, let $H_A$ denote the $A-$ humpty point. Then $H_A$ is the reflection point of $D$ wrt $BC.$ Proof: Let $D'$ be the reflection of $H_A$ wrt $BC.$ Note that $$180-A=\angle BH_AC=\angle BD'C\implies D'\in (ABC)$$$$\angle H_ABC=\angle BCD'=\angle BAD'\implies AD, AH_A \text { isogonal }\implies D'\in \text { symmedian } $$But since $ABDC$ is harmonic, we get that $D$ lies on the symmedian. Hence $D'=D.$ Hence by our Lemma , we get $B'$ as the $A$ humpty point of the triangle. And by our definition of Humpty point, we get that the circle passing through $A,B', D$ touches $AC .$

I found a very long solution (may be difficult to understand) using inversion and moving points...

Corollary of this

Let $M$ be the midpoint of $BD.$ Since $ABCD$ is harmonic $$\measuredangle MAD=\measuredangle BAC=\measuredangle CAB',\text{ } \measuredangle MCD=\measuredangle ACB'.$$Thus $M$ is isogonal conjugate of $B'$ wrt $ABC$, and $\measuredangle ADB'=\measuredangle MDC=\measuredangle BAC=\measuredangle CAB'$ as desired.

so this is mainly about completing well known configurations for instance consider Brazil 2011/5

The condition implies $ABCD$ is harmonic so $BD$ is $B$ symmedian in $\Delta ABC$. This implies $B'$ is the $B$ humpty point in $\Delta BAC$ so this finishes since the required is a well known consequence.