Let $ABCD$ be a cyclic quadrilateral and denote by $\mathcal C$ its circumscribed circle. Denote by $P$ the intersection of the lines $(AB)$ and $(CD)$ ; and denote by $Q$ the intersection of the lines $(BC)$ and $(AD)$. Let $\mathcal C_1$ be the circumscribed circle of the triangle $PQA$, and let $\mathcal C_2$ be the circumscribed circle of the triangle $PQC$.

I would like to prove that if the circles $\mathcal C_1$ and $\mathcal C$ are tangent in point $A$, then the circles $\mathcal C_2$ and $\mathcal C$ are tangent in point $C$. Moreover, I'd like the answer to be as elementary as possible.

If I denote by $O, O_1$ and $O_2$ the centers respectively of the circles $\mathcal C, \mathcal C_1$ and $\mathcal C_2$, then it all boils down to proving that points $O, O_2$ and $C$ are aligned. For this purpose, I have tried computing the angle $(\vec{O_2O},\vec{O_2C})$ using various Chasles relations and inscribed angles theorems, but to no avail.

While looking for a solution, I have noticed that the circles $\mathcal C_1$ and $\mathcal C_2$ have the same radii. Moreover, the quadrilateral $O_1PO_2Q$ is a rhombus. These facts be used freely, it follows essentially from the law of sines.

## Best Answer

Let me see if I remember what I thought ...

So, we consider the inversion defined above, with center at $P$ and radius $\sqrt{PA\cdot PB}$. The radius has been chosen such that $A$ is the inverse of $B$. By power of the point or similarity of triangles $PA\cdot PB=PD\cdot PC$. So $D$ is the inverse of $C$ as well. The circle $\mathcal{C}$ is then its own inverse.

The inverse of the circle $\mathcal{C}_1$, or $APQ$, since it passes through the center of inversion, is the line $BQ'$. Since $\mathcal{C}_1$ is tangent to $\mathcal{C}$ at $B$, then $BQ'$ is tangent to $\mathcal{C}$ at $B$. Recall that inversion preserves the magnitude of angles. In this case angle zero between the two circles.

The inverse of the circle $\mathcal{C}_2$, or $PQC$, is the line $DQ'$. Proving that $\mathcal{C}_2$ is tangent to $\mathcal{C}$ is equivalent to proving that $DQ'$ is tangent to $\mathcal{C}$.

The inverse of the line $ADQ$ is a circle passing through the center of inversion, $P$. Therefore, $PBCQ'$ is a cyclic quadrilateral.

Finally, since $C,D$ are inverses of each other and $Q,Q'$ are inverses of each other, then $\angle Q'DP=\angle CQP$. (You can also prove this by similarity of triangles, since by definition of inversion $PQ\cdot PQ'=PD\cdot PC$).

I think those are the observations that we need from inversion, to then do angle chasing.

Now, $\angle BDC=\angle CBQ'$ for being inscribed on the same arc. [I am not sure if $\angle CBQ'$ is called inscribed or if a more specific names is reserved for it]. Since $PBCQ'$ is cyclic, then $\angle CBQ'=\angle CPQ'$. Therefore, $BD|| PQ$. Since these two are parallel, then $\angle PQB=\angle DBQ$.

The proof end now recalling that $\angle PQB=\angle PDQ'$. Therefore, $\angle DBC=\angle CDQ'$, which implies that $DQ'$ is tangent to $\mathcal{C}$ at $D$.

The use of inversion can be removed, of course. It is just that it is the first thing that crossed my mind. My suggestion would be to try to get first that $BD||PQ$. This definitely follows from $\mathcal{C}_1$ being tangent to $\mathcal{C}$, just using proportions (power of the points $P$ and $Q$ with respect to $\mathcal{C}$). Then from that get to what is wanted chasing angles. Perhaps defining $Q'$ and the intersection of the circle $PBC$ with $PQ$ is useful for this last step.