Here is an image,that i've promised to upload earlier. Image not found

Remark: The condition $ A,D,B$ are collinear is not necessary.I did not use it in my proof. I'm writing this,because it make the problem more beautiful,and kinda easier,for someone,who is trying to solve the problem with using all the conditions. Best regards.

$ OD\perp MH$ equivalent$ {OM}^2-{OH}^2={DM}^2-{OH}^2$......

What is it?If you really solved it,please post your solution,if you don't,then there is no reason,to post something like a hint,which is not true

Erken wrote: Let $ C$ and $ D$ be two intersection points of circle $ O_1$ and circle $ O_2$.A line,passing through $ D$,intersects with circle $ O_1$ and circle $ O_2$ at points $ A$ and $ B$ respectively.The points $ P$ and $ Q$ are on circles $ O_1$ and $ O_2$ respectively.The lines $ PD$ and $ AC$ intersect at $ H$,and the lines $ QD$ and $ BC$ intersect at $ M$.Suppose that $ O$ is the circumcenter of the triangle $ ABC$,prove that $ OD\perp MH$ is and only if $ P,Q,M$ and $ H$ are concyclic. I'll add an image later,because there was an image on the problem list. Nice but easy! We have: $ P,Q,M,H$ are concylic $ \Leftrightarrow DP.DH=DM.DQ \Leftrightarrow HD.HP-HD^2=MD.MQ-MD^2$ $ \Leftrightarrow DM^2-DH^2=MD.MQ-HP.HD=MB.MC-HB.HA=(OM^2-R^2)-(OH^2-R^2)=OM^2-OH^2 \Leftrightarrow OD \perp MH$

Hey Erken my solution is the same 10maths_tp0609

dlt5 wrote: Hey Erken my solution is the same 10maths_tp0609 It is my proof too. Yes,indeed the problem is very easy,but nice.

Quote: We have: $ P,Q,M,H$ are concylic $ \Leftrightarrow DP.DH = DM.DQ \Leftrightarrow HD.HP - HD^2 = MD.MQ - MD^2$ $ MD.MQ - HP.HD = MB.MC - HB.HA$ Very nice solution but I think $ MD.MQ-HP.HD=MB.MC-HA.HC$

Different solution. Indeed $A-D-B$ is not required. Let $T_1$ and $T_2$ be the circumcircles of $ABC$ and $PQD$. Let $O$ and $C_2$ be their centers, respectively. I use a well known lemma: If $ABC$ is a triangle, $O$ its circumcenter, $X$ and $Y$ points in $AB$ and $AC$ then $BXYC$ is cyclic if and only if $AO \perp XY$. This is easy to prove. From $HP \cdot DD = HA \cdot HC$ we get $H$ is in the radical axis of $T_1$ and $T_2$. Similarly $M$ is also on this radical axis. Hence we have $MH \perp C_1C_2$. If $PHMQ$ is cyclic then $C_2D \perp MH$ by the lemma which implies $O - C_2 - D$ and $OD \perp MH$. SImilarly if $OD \perp MH$ we have $D-O-C_2$ so $DC_2\perp MH$ so by lemma $PHMQ$ is cylic and we are done.