Given is a parallelogram $ABCD$ with $\angle A < 90^o$ and $|AB| < |BC|$. The angular bisector of angle $A$ intersects side $BC$ in $M$ and intersects the extension of $DC$ in $N$. Point $O$ is the centre of the circle through $M, C$, and $N$. Prove that $\angle OBC = \angle ODC$. [asy][asy] unitsize (1.2 cm); pair A, B, C, D, M, N, O; A = (0,0); B = (2,0); D = (1,3); C = B + D - A; M = extension(A, incenter(A,B,D), B, C); N = extension(A, incenter(A,B,D), D, C); O = circumcenter(C,M,N); draw(D--A--B--C); draw(interp(D,N,-0.1)--interp(D,N,1.1)); draw(A--interp(A,N,1.1)); draw(circumcircle(M,C,N)); label("$\circ$", A + (0.45,0.15)); label("$\circ$", A + (0.25,0.35)); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$M$", M, SE); dot("$N$", N, dir(90)); dot("$O$", O, SE); [/asy][/asy]
Problem
Source: Dutch NMO 2020 p3
Tags: geometry, parallelogram, equal angles
23.11.2020 20:48
Pretty straight-forward. Obviously, $\triangle ADN$ and $\triangle ABM$ are isosceles. Thus, $OC\perp MN$ and we get that $\angle CMO=\angle MCO = \angle OCN$. Thus, $$\angle BMO=180^{\circ}-\angle CMO=180^{\circ}-\angle MCO=180^{\circ}-\angle MCO-\angle OCN+\angle MCO=\angle DCB+\angle MCO=\angle DCO$$Also, by isosceles triangles, we have that $DC=AB=BM$ and $OM=OC$, hence we conclude that $\triangle DCO\sim \triangle BMO$. Thus, $\angle MOB=\angle DOC\implies \angle DOB=\angle MOC=180^{\circ}-\angle MCO-\angle OCN=\angle DCB$, hence $DCOB$ is cyclic, which means that $\angle OBC = \angle ODC$.
23.11.2020 20:50
rafaello wrote: Pretty straight-forward. Obviously, $\triangle ADN$ and $\triangle ABM$ are isosceles. Thus, $OC\perp MN$ and we get that $\angle CMO=\angle MCO = \angle OCN$. Thus, $$\angle BMO=180^{\circ}-\angle CMO=180^{\circ}-\angle MCO=180^{\circ}-\angle MCO-\angle OCN+\angle MCO=\angle DCB+\angle MCO=\angle DCO$$Also, by isosceles triangles, we have that $DC=AB=BM$ and $OM=OC$, hence we conclude that $\triangle DCO\sim \triangle BMO$. Thus, $\angle MOB=\angle DOC\implies \angle DOB=\angle MOC=180^{\circ}-\angle MCO-\angle OCN=\angle DCB$, hence $DCOB$ is cyclic, which means that $\angle OBC = \angle ODC$. You can also finish by spiral similarity, since we see that $B=DB\cap CM$, spiral similarity centre is $O$ and it takes triangle $DCO$ to $BMO$, thus circumcircle of $(DCO)$ passes through $B$.
17.01.2022 08:32
DC = AB = BM and OC = OM and ∠OMB = 180 - ∠OMC = 180 - ∠ONC = 180 - ∠OCN = ∠OCD so OMB and OCD are congruent so ∠OBC = ODC.
03.10.2024 20:56
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.82, xmax = 19.82, ymin = -7.48, ymax = 10.36; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen ffqqff = rgb(1,0,1); pen ffqqtt = rgb(1,0,0.2); pen qqccqq = rgb(0,0.8,0); pen ffwwqq = rgb(1,0.4,0); pen wwqqcc = rgb(0.4,0,0.8); pen ffdxqq = rgb(1,0.8431372549019608,0); draw((-5,-3)--(1,-3)--(4,7)--(-2,7)--cycle, linewidth(0.7) + zzttqq); draw(arc((-5,-3),0.6,36.65037788300319,73.30075576600639)--(-5,-3)--cycle, linewidth(0.7) + ffqqtt); draw(arc((-5,-3),0.6,0,36.65037788300319)--(-5,-3)--cycle, linewidth(0.7) + ffqqtt); /* draw figures */ draw((-5,-3)--(1,-3), linewidth(0.7) + blue); draw((1,-3)--(4,7), linewidth(0.7) + blue); draw((4,7)--(-2,7), linewidth(0.7) + blue); draw((-2,7)--(-5,-3), linewidth(0.7) + blue); draw((xmin, 0.744030650891055*xmin + 0.7201532544552742)--(xmax, 0.744030650891055*xmax + 0.7201532544552742), linewidth(0.7) + dotted + white); /* line */ draw((4,7)--(8.44340443311505,7.002304950562283), linewidth(0.7) + ffqqff); draw(circle((6.2232511786597735,4.015116599075233), 3.721877848684488), linewidth(0.7) + ffqqtt); draw((1,-3)--(8.44340443311505,7.002304950562283), linewidth(0.7) + qqccqq); draw((-2,7)--(1,-3), linewidth(0.7) + ffwwqq); draw((-2,7)--(6.2232511786597735,4.015116599075233), linewidth(0.7) + wwqqcc); draw((2.7240873133980723,2.746957711326907)--(6.2232511786597735,4.015116599075233), linewidth(0.7) + ffdxqq); draw((4,7)--(6.2232511786597735,4.015116599075233), linewidth(0.7) + ffdxqq); draw((-5,-3)--(8.44340443311505,7.002304950562283), linewidth(0.7) + ffwwqq); /* dots and labels */ dot((-5,-3),dotstyle); label("$A$", (-5.14,-3.46), NE * labelscalefactor); dot((1,-3),dotstyle); label("$B$", (0.94,-3.56), NE * labelscalefactor); dot((4,7),dotstyle); label("$C$", (3.9,7.26), NE * labelscalefactor); dot((-2,7),dotstyle); label("$D$", (-1.92,7.2), NE * labelscalefactor); dot((8.44340443311505,7.002304950562283),dotstyle); label("$N$", (8.46,7.24), NE * labelscalefactor); dot((2.7240873133980723,2.746957711326907),linewidth(4pt) + dotstyle); label("$M$", (2.64,2.02), NE * labelscalefactor); dot((6.2232511786597735,4.015116599075233),linewidth(4pt) + dotstyle); label("$O$", (6.3,3.66), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] A synthyetic Geo solution same as @above. $\color{red}\textbf{Claim:-}$ $\angle OBC=\angle ODC$ or We have to prove that the two triangles are congruent. That is Triangles $OBM \cong ODC$ $\color{blue}\textbf{Proof:-}$ First Join $O,M.$ Now we get the points $O, M, C, N$ are concyclic. Now we also get some angle conditions, $$\angle BAN=\angle NAD=\angle BMA=\angle MNC=\angle NMC=2\angle BAD=2\angle BCD=2\angle NOC$$Now we get that $DC=AB=BM$ and $OC=OM.$ Some other angle conditions are $$\angle OCD=\angle OMB=180-\angle OMC=180-\angle ONC=180-\angle OCN.$$From above angle conditions we get that the triangles $$OMB \cong OCD \implies \boxed{\angle OBC=\angle ODC.}$$