We will first solve problem b), then problem a). The points D, E, F lie on the angle bisectors of triangle ABC, i. e. on the lines AI, BI, CI, where I is the center of the incircle of triangle ABC. [This is because the angle bisector of an angle in a triangle always passes through the midpoint of the arc cut off from the circumcircle by the opposite side.] Now, we consider the triangle DEF. We have < DIB = 180° - < AIB = < IAB + < IBA (sum of angles in triangle AIB) = A/2 + B/2, while < IDB = < ADB = < ACB (cyclic) = C = 180° - A - B = 180° - 2 (A/2 + B/2) = 180° - 2 < DIB. Thus, < DBI = 180° - < DIB - < IDB = 180° - < DIB - (180° - 2 < DIB) = - < DIB + 2 < DIB = < DIB, so that triangle BDI is isosceles, i. e. we have BD = ID. Similarly, CD = ID. Hence, BD = ID = CD, and the point D is the circumcenter of triangle BIC. Similarly, E and F are the circumcenters of triangles CIA and AIB. [This is an important fact: that the midpoint of the arc BC on the circumcircle of triangle ABC is the circumcenter of triangle BIC.] Now, since the points E and F are the circumcenters of triangles CIA and AIB, both of them lie on the perpendicular bisector of the segment AI. Hence, the line EF is the perpendicular bisector of the segment AI. Similarly, the lines FD and DE are the perpendicular bisectors of the segments BI and CI. In other words, the lines EF, FD, DE pass through the midpoints of the segments AI, BI, CI, respectively, and are perpendicular to these segments. Hence, the point P, being defined as the point of intersection of EF and AI (you have said AD, but this is the same line), must be the midpoint of AI. Now, since DE is perpendicular to CI, the line CI is an altitude of triangle GCH. On the other hand, it is obviously an angle bisector of this triangle. But if an altitude and an angle bisector of a triangle coincide, then the triangle is isosceles; hence, the triangle GCH is isoscles, and the midpoint M of its base GH must lie on the altitude from the apex C. In other words, the point M lies on the line CI. But since the point M also lies on the line DE, and the line DE is the perpendicular bisector of CI, the point M must be the midpoint of CI. Similarly, N is the midpoint of BI. Thus, we have found: The points P, N, M are the midpoints of the segments AI, BI, CI, respectively. Now, since N and M are the midpoints of the sides BI and CI of triangle BIC, we have MN || BC. So we have proven that the line MN is parallel to the line BC. This is the first important assertion. Now we will show that the circumcenter of triangle DMN lies on the circumcircle of triangle PMN. Since the lines EF, FD, DE are perpendicular to the lines AI, BI, CI, respectively, the lines AI, BI, CI are the three altitudes of the triangle DEF. Consequently, the point I is the orthocenter of this triangle, and the points P, N and M are the feet of the altitudes of this triangle. Hence, the circumcircle of triangle PMN is the nine-point circle of triangle DEF. Therefore, it remains to show that the circumcenter of triangle DMN lies on the nine-point circle of triangle DEF. But since < DNI = 90° and < DMI = 90°, the points N and M lie on the circle with diameter DI; hence, the circumenter of triangle DMN is the center of the circle with diameter DI, i. e. the midpoint of the segment DI, hence the midpoint of the segment joining a vertex of triangle DEF with its orthocenter. But it is well-known that such midpoints always lie on the nine-point circle. Hence, the circumenter of triangle DMN lies on the nine-point circle of triangle DEF, i. e. on the circumcircle of triangle PMN. Thus, problem b) is solved. Now, problem a) is simple: Since MN || BC, we have < DMN = < DGK. Now, since the line DE is the perpendicular bisector of the segment CI, and the point M is the midpoint of the segment CI, the triangle CMG is right-angled at M, so we have < CGM = 90° - < GCM. Finally, < GCM = < BCI = C/2. Thus, < DMN = < DGK = < CGM = 90° - < GCM = 90° - C/2. Similarly, < DNM = 90° - B/2. Thus, by the sum of the angles in triangle DMN, we have < MDN = 180° - < DMN - < DNM = 180° - (90° - C/2) - (90° - B/2) = (B + C) / 2 = (180° - A) / 2 (sum of angles in triangle ABC) = 90° - A/2. Hence, we found all three angles of triangle DMN. Problem a) is solved. Darij

is a) = 60 ? can anyone draw a diagram plz

Let $FC$, $EB$ intersect $DE$, $FD$ at $M'$, $N'$ respectively. We will prove first that $M' = M, N = N'$ and that lines $AD$, $BE$, $FC$ are altitudes of the $\triangle DEF$. It's easy to see that lines $AD$, $BE$ and $CF$ form the internal angle bisectors of $\triangle ABC$. Consequently, we can determine the $\angle FDE$ of $\triangle DEF$ as being equal to $\angle FDA + \angle ADE = \angle C/2 + \angle B/2 = 90^{\circ} - \angle A/2$ Also we have $\angle DFC = \angle A/2$, thus $\angle FM'D = 90^{\circ}$. Similarly $\angle EN'D = 90^{\circ}$. Thus $AD$, $BE$, $FC$ are altitudes of the $\triangle DEF$ with $P$, $N'$, $M'$ respectively being the feet of the altitudes. Now since $M'C$ is internal bisector of $\angle HCG$ and $CM'$ is perpendicular to $GH$, we have that $CM'$ is the perpendicular bisector of $GH$. Hence $M' = M$. Similarly it can be shown that $BN'$ is the perpendicular bisector of $KJ$, and hence $N' = N$. Now lines $AD$, $BE$ and $CF$ intersect at point $Q$. So $Q$ is the incenter of $\triangle ABC$ and orthocenter of $\triangle DEF$. Clearly, $QNDM$ is a cyclic quadrilateral as $N$, $M$ are the feet of perpendiculars from $E$ and $F$. So, we have $\angle QNM = \angle QDM = \angle ADE = \angle B/2$. Similarly, since $PQNF$ is also a cyclic-quadrilateral, reasoning as above, $\angle QNP = \angle PFQ = \angle CFE = \angle B/2$. Thus we have that $\angle QNM = \angle QNP$ and so $NQ$ is an internal bisector of $\angle PNM$. Reasoning in a similar fashion it can be proven that $PQ$ and $MQ$ are internal bisectors of other 2 angles of $\triangle PNM$. Thus $Q$ also happens to be the incenter of $\triangle PNM$ in addition to being that of $\triangle ABC$. $Part 1$: Angles of $\triangle DMN$: Since $\angle QNM = \angle B/2, \angle DNM = 90^{\circ} - \angle B/2$. Similarly $\angle NMD = 90^{\circ} - \angle C/2$. Finally $\angle NDM = 90^{\circ} - \angle A/2$. $Part 2$: Let circumcircle of $\triangle PMN$ cut line $DQ$ at point $R$. Since $PMRN$ is a cyclic quadrilateral, we have $\angle RNM = \angle RPM = \angle A/2$. Similarly, $\angle RMN = \angle RPN = \angle A/2$. Thus $RN$ = $RM$. Now, $\angle RND = 90^{\circ} - (\angle A/2 + \angle B/2) = \angle C/2$ and $\angle RDN = \angle ADF = \angle C/2$. Thus $RN$ = $RD$. Thus we have $RN$ = $RM$ = $RD$. So $R$ is the circumcenter of $\triangle DMN$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(22cm); 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 = 3.9048964909723813, xmax = 28.37872103143898, ymin = -5.205353747461378, ymax = 6.179013037208239; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pair A = (12.33196188674394,5.090785052545904), B = (10.471439999121461,-2.4835346767002537), C = (19.53327218330795,-2.50750777771662), D = (14.995812108182063,-4.96914681354665), F = (9.779110881444309,1.702190924774591), G = (16.062168836308647,-2.49832496462932), H = (17.14551955555843,0.011873130022662116), M = (16.60384419593354,-1.243225917303329), K = (13.057519304633773,-2.490376156344466), J = (11.088336903046809,0.027897422754177045), P = (13.003189047651531,2.555920497827937), Q = (13.674416208559121,0.021055943109969937); draw(A--B--C--cycle, linewidth(2.) + zzttqq); /* draw figures */ draw(A--B, linewidth(0.8) + zzttqq); draw(B--C, linewidth(0.8) + zzttqq); draw(C--A, linewidth(0.8) + zzttqq); draw(circle((15.010061924088253,0.41728359899296524), 5.386449261469474), linewidth(0.8)); draw((xmin, 0.9477537201360008*xmin-13.808558429033418)--(xmax, 0.9477537201360008*xmax-13.808558429033418), linewidth(0.4) + dotted); /* line */ draw((xmin, -0.24563550973938744*xmin + 4.104287810936154)--(xmax, -0.24563550973938744*xmax + 4.104287810936154), linewidth(0.4) + dotted); /* line */ draw((xmin, 378.000000000003*xmin-5673.386123706412)--(xmax, 378.000000000003*xmax-5673.386123706412), linewidth(0.4) + dotted); /* line */ draw(D--(18.91961706710927,4.122579029867964), linewidth(0.8)); draw(F--D, linewidth(0.8)); draw((12.07292810384029,-1.2312393667951447)--M, linewidth(0.8)); draw(A--D, linewidth(0.8)); draw(F--(18.91961706710927,4.122579029867964), linewidth(0.8)); draw(circle((14.335114158370592,-2.4740454352183394), 2.581095246018813), linewidth(0.8)); draw(circle((14.342239066323689,0.21916977105146512), 2.69322463073474), linewidth(0.8)); draw(C--F, linewidth(0.8)); draw(B--(18.91961706710927,4.122579029867964), linewidth(0.8)); draw(M--P, linewidth(0.8)); draw(P--(12.07292810384029,-1.2312393667951447), linewidth(0.8)); /* dots and labels */ dot(A,dotstyle); label("$A$", (12.032924127774804,5.3585181338086265), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (10.19963082799129,-2.5643857771438765), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (19.712243614280563,-2.5900262428751146), NE * labelscalefactor); dot(B,linewidth(4.pt) + dotstyle); dot((19.533272183307947,-2.5075077777166195),linewidth(4.pt) + dotstyle); dot((19.53327218330795,-2.5075077777166195),linewidth(4.pt) + dotstyle); dot((18.91961706710927,4.122579029867964),linewidth(4.pt) + dotstyle); label("$E$", (19.10969266959647,4.0893150801123515), NE * labelscalefactor); dot(D,linewidth(4.pt) + dotstyle); label("$D$", (15.020038385464021,-5.166893048864521), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (9.494518020382248,1.8201338628977997), NE * labelscalefactor); dot(G,linewidth(4.pt) + dotstyle); label("$G$", (16.237960507697824,-2.8079702015906367), NE * labelscalefactor); dot(H,linewidth(4.pt) + dotstyle); label("$H$", (17.39178146560353,0.025301261711148514), NE * labelscalefactor); dot(M,linewidth(4.pt) + dotstyle); label("$M$", (16.91743284957563,-1.1926208605226505), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("$K$", (12.699576236786989,-2.8336106673218744), NE * labelscalefactor); dot(J,linewidth(4.pt) + dotstyle); label("$J$", (10.789361539809764,-3.392040200893613e-4), NE * labelscalefactor); dot((12.07292810384029,-1.2312393667951447),linewidth(4.pt) + dotstyle); label("$N$", (11.661137374671853,-1.2695422577163642), NE * labelscalefactor); dot(P,linewidth(4.pt) + dotstyle); label("$P$", (13.007261825561844,2.8585727250129334), NE * labelscalefactor); dot(Q,linewidth(4.pt) + dotstyle); label("$Q$", (13.725194866036507,0.3073463847547651), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Sorry for the bad diagram so here the main thing is to notice is that we can see that $DFAC$ is cyclic quadrilatral so we can say that $\angle FDA=\angle FCA=C/2$ Similarly, $\angle ADE=\angle CFE=B/2$ $\angle BEF=C/2$ $\angle DFC=\angle DEB=A/2$ Claim: $AD,CF,BE$ are the altitudes of the triangle $DEF$ Proof: so we can see here that $\angle FDE=\angle FDA+\angle ADE=C/2+B/2=90^0-A/2$ and we know that $\angle DFC=A/2$ since by interior angle property we can say that $\angle FDE+\angle DFC=90-A/2+A/2=90^0$ since $CF$ is perpendicular altitudes of $\triangle DEF$ similarly $AD \text{ and } BE $ are the altitudes of the $\triangle DEF$ $\square$ Let the altitudes intersect at $Q$ so now join $P,M,N$ and we can see that $PMN$ is orthic triangle of $\triangle DEF$ now we can see that $PQME$ is cyclic quadrilatral so we can say that \[\angle QEP=\angle PMQ=C/2\]\[\angle MEQ=\angle MPQ=A/2\]similarly $PQNF$ is cyclic quadrilatral so applies that\[\angle PFQ=\angle PNQ=B/2\]and since we know that triangle made by the perpendiculars of the triangle is orthocenter is the incircle of the formed triangle so we can say that \[\angle MPD=NPD=A/2\]\[\angle MNE=\angle PNE=B/2\]\[\angle PMF=\angle NMF=A/2\]\textit{Part 1} \[\angle NDM=B/2+C/2 =90^0-A/2\]\[\angle DMN=90^0-C/2\]\[\angle DNM=90-B/2\]and we are done with part 1 $\square$ \textit{Part 2} Let suppose the circumcircle of $\triangle PMN$ cuts $DQ$ at $Q_1$ now its obvious that $PMQ_1N$ is cyclic quadrilatral so we can see that \[\angle Q_1NM=\angle Q_1PM=A/2\]\[\angle Q_1MN=\angle Q_1PN=A/2\]so since we can see that $\angle Q_1NM=\angle Q_1MN$ so we can say that $Q_N=Q_M$ now see in triangle $DNM$ we have \[\angle DNQ_1=90-A/2-B/2=C/2=\angle NDQ\]so by this we can say that $Q_1N=Q_1D$ since we have $Q_1N=Q_1M=Q_1D$ so we can say that $Q_1$ is the circumcenter of triangle $DMN$ and we are done with the both part $\blacksquare$

Equal arcs = Cyclic Quads = Problem is solved ! $I$ is the incenter of $\triangle ABC$ & $O$ is the circumcenter of $\triangle DMN$ and let $2A,2B,2C$ denote $\angle BAC,\angle ABC, \angle ACB$ I will divide the proof into 2 parts

It is far more easier to solve the problem than to latex it Diagram for reference