The circles $\odot (ABCD),\ \odot (AOC)$ are orthogonal and the tangents at $A,\ C$ concur on $BD$ iff the tangents at $B,\ D$ concur on $AC$, i.e. $ABCD$ is harmonic. Best regards, sunken rock

sunken rock wrote: The circles $\odot (ABCD),\ \odot (AOC)$ are orthogonal and the tangents at $A,\ C$ concur on $BD$ iff the tangents at $B,\ D$ concur on $AC$, i.e. $ABCD$ is harmonic. Best regards, sunken rock Sorry dear Sunken Rock , but $ABCD$ is not necessary harmonic .

mahanmath wrote: The quadrilateral $ABCD$ is inscribed in a circle with center $O$. The diagonals $AC$ and $BD$ do not pass through $O$. If the circumcentre of triangle $AOC$ lies on the line $BD$, prove that the circumcentre of triangle $BOD$ lies on the line $AC$. $P, Q$ are circumcenters of $\triangle AOC, \triangle BOD.$ Inversion in $(O)$ takes $\odot (AOC)$ to $AC$ and $BD$ to $\odot(BOD).$ $P \in BD$ $\Longleftrightarrow$ $\odot (AOC) \perp BD$ $\Longleftrightarrow$ $AC \perp \odot(BOD)$ $\Longleftrightarrow$ $Q \in AC.$

yetti wrote: mahanmath wrote: The quadrilateral $ABCD$ is inscribed in a circle with center $O$. The diagonals $AC$ and $BD$ do not pass through $O$. If the circumcentre of triangle $AOC$ lies on the line $BD$, prove that the circumcentre of triangle $BOD$ lies on the line $AC$. $P, Q$ are circumcenters of $\triangle AOC, \triangle BOD.$ Inversion in $(O)$ takes $\odot (AOC)$ to $AC$ and $BD$ to $\odot(BOD).$ $P \in BD$ $\Longleftrightarrow$ $\odot (AOC) \perp BD$ $\Longleftrightarrow$ $AC \perp \odot(BOD)$ $\Longleftrightarrow$ $Q \in AC.$ Thanks dear yetti ! This is also my solution .Actually for one who knows inversion , everything in problem push him\her to use it , but what for the others ? I have a good news for them !! , this problem has a solution which only use Pythagoras .

Let $E \equiv AC \cap BD.$ Power of $E$ to $(O)$ is $EA \cdot EC = EB \cdot ED$ $\Longrightarrow$ $OE \perp PQ$ is radical axis of $(P) \equiv \odot(AOC)$ and $(Q) \equiv \odot(BOD).$ $P \in BD$ $\Longleftrightarrow$ $PE \equiv BD$ $\Longleftrightarrow$ $PE \perp OQ$ $\Longleftrightarrow$ $E$ is orthocenter of $\triangle OPQ$ $\Longleftrightarrow$ $QE \perp OP$ $\Longleftrightarrow$ $QE \equiv AC$ $\Longleftrightarrow $ $Q \in AC.$

yetti wrote: Let $E \equiv AC \cap BD.$ Power of $E$ to $(O)$ is $EA \cdot EC = EB \cdot ED$ $\Longrightarrow$ $OE \perp PQ$ is radical axis of $(P) \equiv \odot(AOC)$ and $(Q) \equiv \odot(BOD).$ $P \in BD$ $\Longleftrightarrow$ $PE \equiv BD$ $\Longleftrightarrow$ $PE \perp OQ$ $\Longleftrightarrow$ $E$ is orthocenter of $\triangle OPQ$ $\Longleftrightarrow$ $QE \perp OP$ $\Longleftrightarrow$ $QE \equiv AC$ $\Longleftrightarrow $ $Q \in AC.$ I think " $P \in BD$ $\Longleftrightarrow$ $PE \equiv BD$ " is not right. Edit: I misread PE=BD. The proof is right .

let arc $AC=\alpha$,arc $BD=\beta$,the acute angle between $AC$ and $BD$ be $\omega$ then the two conclusions are all equivalent to $2cos\alpha cos\beta=sin\omega$

I also used inversion but concluded it in a little different way.... Circumcenter of $\triangle AOC$ goes to the point $O_1$ which is the reflection of $O$ wrt $AC$. Since $BD$ passes through the circumcenter of $AOC\Longrightarrow B,D,O_1,O$ are cyclic.So centre of $BOD$ lies on the perpendicular bisector of $OO_1$ which is nothing but $AC$.So done.

Ok, by pythegoras Let $ABCD$ be inscribed in $\Gamma$ . Let $T$ be the center of the circumcircle $\omega_1$ of $\triangle AOC$ which lies on the line $BD$. Let $R$ be the center of the circumcircle $\omega_2$ of $\triangle BOD $ The power of $R$ wrt $\Gamma$ is $|OR^2 - OB^2|= |RB^2 - OB^2 |= |RL^2 + BL^2 - OL^2 - BL^2 | = | RL^2 - OL^2|$. The power of $R$ wrt $\omega_1$ is $|RT^2 - OT^2| = |RL^2 + LT^2 - OL^2 - LT^2|= |RL^2 - OL^2|$ So $R$ lies on the radical axis of $\Gamma$ and $\omega_1$, that is $AC$.

WLOG, suppose the circumcircle is the unit circle centred at origin. then the circumcentre of ttriangle $AOC$ is $\frac{ac}{a+c}$ and that of triangle $BOD$ is $\frac{bd}{b+d}$. then after simplification implies that , $ac+bd= (a+c)(b+d) \implies \frac{bd/[b+d]-c}{1/[b+d]-1/c}=-ac$ and we are done

Use the following Lemma: If $P$ is a fixed point, and $C$ a circle with $O$ it's center. Then if we have a non-fixed line $l$;$P\in l$, $l \cap C=\{B,D\}$, then the locus of the center of the circle $(OBD)$ lies on a line.

Let $P,Q$ be the circumcenters of triangles $AOC$ and $BOD$ and $R$ be the reflection of $O$ in line $BD$. Point $P$ lies on line $BD$ then $PR=PO$, so $R$ lies on the circle $(AOC)$. The inversion about circle $(ABCD)$ sends $R$ to $Q$ and circle $(AOC)$ to line $AC$ so $Q$ lies on the line $AC$. $\, \square$