parmenides51 wrote: Let $ABCD$ be a parallelogram such that $|AB| > |BC|$. Let $O$ be a point on the line $CD$ such that $|OB| = |OD|$. Let $\omega$ be a circle with center $O$ and radius $|OC|$. If $T$ is the second intersection of ω and $CD$, prove that $AT, BO$ and $\omega$ are concurrent. Proposed by Ivan Novak My Solution. Let $AT \cap OB = X.$ We aim to show $X \in (\omega).$ Given, $CD \cap \omega = T (\ne C)$, we further let $\overline{OB} \cap (\omega) = M,$ $L$ and $Y$ denote the foot from $O$ onto $\overline{TM}$ and $\overline{BD}$ respectively. As $OB=OD, OT=OM$, so $TM \parallel BD,$ and thus $\angle OBD = \angle ODB = \alpha - (*)$ (say), hence $$\alpha = \angle OTM = \angle OTM.$$Next, we observe that $AY=YC$, and again as $OT=OC$, so $OY \parallel AT$. These two parallel conditions along with the $90^{\circ}$s at $Y,L$, gives $TKYL$ as a rectangle, where $K = \overline{AT} \cap \overline{BD}.$ So, we have $$\angle ATM = 90^{\circ} = \angle XTM.$$Now, from $(*)$, we get $$\angle MOC = 2\alpha \implies \angle OCM = 90^{\circ} - \alpha.$$Since, $\triangle TXM,$ we had $\angle XTM = 90^{\circ}$, and also $\angle OMT = \angle XMT = \alpha$, whence $$\angle TXM = 90^{\circ} - \alpha = \angle OCM = \angle TCM,$$with $T,M,C$ on $(\omega)$, ergo $X$ lies on $(\omega),$ and $AT, BO$ and $\omega$ are concurrent. $\blacksquare$

Solution. Let $AT \cap BO = X$. Let $DT=b,OT=OC=a$. Obviously, $\triangle ABX \sim \triangle TOX$. Thus $$\frac{OX}{OB}=\frac{OT}{AB-OT}$$$$\frac{OX}{a+b}=\frac{a}{a+b}$$Thus $OX=OT=OC=a \implies X \in \omega$.

Solution. Let $AT \cap \omega = X$, we need to show that $\overline{A-T-X}$ are collinear. Now let $AC \cap BD= Y$, $OY\perp BC\implies \angle YOD=90^{\circ}-\angle BDC.$ $$\angle ODB=\angle OBD=\angle ABD,\angle DOX=\angle BOC=\angle ABO=2\angle ABD\implies \angle OTX=\angle OXT=90-\angle BDC.$$Now we get that $\angle YOD=\angle YOT=\angle OTX\implies TX\parallel YO$. Also, it is clear that $YO\parallel AT$. $TX\parallel YO$, and $YO\parallel AT\implies \overline{A-T-X}$ are collinear.$\blacksquare$

Let $S$ be the point on $BO$, such that $AB=SB$, then $$OS=SB-OB=AB-OD=CD-OD=OC,$$thus $S$ lies on $\omega$. Now let us show that $A,T,S$ are collinear and we are done. This is true, since if $T'=AS\cap CD$, then $$\measuredangle T'SO=\measuredangle ASB=\measuredangle BAS=\measuredangle BAT'=\measuredangle DT'A=\measuredangle OT'S\implies OT'=OS=OC,$$hence $T'$ lies $\omega$. Meaning, $T=T'$, therefore we are done.

Let $\omega\cap OB=E$ $\implies$ $BE=OE+OB=DO+OC=DC=AB$ $\implies$ $\frac{EO}{TO}=\frac{EB}{AB}$. Since $AB||TO$ $\implies$ $A-T-E$ are collinear. So,done.

very nice!

Let AT and Bo meet at S. we will show that OS = OC. OT/AB = SO/SB ---> OT/OC+OD = SO/SO+OB ---> OT/OT+OD = SO/SO+OB , OB = OD ---> OS = OT = OC.

Let $AT \cap BO=X$. Draw a parallel line from $O$ to $AT$ and let it cut $AB$ at $O'$. Notice that $BO'=BO$ and we are done by angle chasing.

Let $BO$ meet $\omega$ at $S$ Claim1 : $BD || CS$. Proof : $\angle DCS = \angle DOS/2 = \angle BDO = \angle BDC$. Note that $TS \perp SC$ so $TS \perp BD$ so for proving $A,T,S$ are collinear we can instead prove $AT \perp BD$. Claim2 : $AT \perp BD$. Proof : Let $AC$ meet $BD$ at $K$. $K$ is midpoint of $BD$ so $OK \perp BD$. we know $AK = KC$ and $TO = OC$ so by Thales Theorem we have $KO || AT$ so $AT \perp BD$. we're Done.

It's easy Let $BO\cap AT =E$ , let $G$ the midpoint of $CD$ claim 1: $AT \parallel OG$ proof: $\frac{CO}{CT}=\frac{CG}{CA}=\frac{1}{2}$ Claim 2: $BD$ is the perpendicular bisector of $AE$. proof:$\angle ODB= \angle OBD = \angle ABD$ and $AE \perp BD$ since $AT \parallel OG $ Claim 3: $OT=OE= OC$ proof: $\angle AEB =\angle TAB =\angle DTA = \angle ETO $. and we are done