First of all, it's easy to see that $AD\perp EF,GH$. The triangles $ACF,ABE$ are isosceles, and have the same angles, so they are similar. This means that $AFE$ is similar to $ACB$, and thus to $AGH$. Since $AFE,AGH$ share the same altitude starting from $A$, it means that $AFE$ is obtained from $AGH$ by reflecting it through a line $\perp AD$, and then performing a homothety centered at $A$ of ratio $\frac{AF}{AG}$, so $EFGH$ will be a rectangle iff $AF=GH$, i.e. iff $AF=AC\iff AFC$ is equilateral. With the given conditions, this is equivalent to $\angle ADC=30^{\circ}$, and the conclusion follows easily.

This problem was problem 2 in the 3rd German TST 2005. Hence, I have a conjecture on where it is from, although some people repeatedly dispute it . Anyway, here is the way the problem was posed on our TST: Problem. Let ABC be an acute-angled triangle with A < B, and let U be the circumcenter of the triangle ABC. The lines CU and AB intersect at a point D. Let E and F be the circumcenters of triangles ACD and BCD. Choose points K and L on the rays AC and BC such that AK = BL = a + b. Prove that the quadrilateral EFKL is a rectangle if and only if B - A = 60°. And here is the solution I gave on the exam (just copied from my writeup, not simplified, hence it will be far more complicated than necessary): We will use non-directed angles. See the accompanying sketch for the arrangement of the points. Since U is the circumcenter of triangle ABC, the central angle theorem yields < BUC = 2 < BAC = 2A. On the other hand, BU = CU, again because U is the circumcenter of triangle ABC. Hence, the triangle BUC is isosceles, and its base angle is therefore $\measuredangle UCB=\frac{180^{\circ}-\measuredangle BUC}{2}=\frac{180^{\circ}-2A}{2}=90^{\circ}-A$. Analogously, < UCA = 90° - B. Similarly, for the circumcenters E and F of triangles ACD and BCD we can find < ECA = < CDA - 90°; < EAC = < CDA - 90°; < FBC = 90° - < CDB; < FCB = 90° - < CDB. Hereby, in the proofs of the equations < ECA = < CDA - 90° and < EAC = < CDA - 90°, we have to apply the central angle theorem in the form < CEA = 2 $\cdot$ acute chordal angle of the chord CA in the circumcircle of triangle ACD = 2 $\cdot$ (180° - < CDA) = 360° - 2 < CDA, since the angle < CDA is obtuse, and the point E lies outside of the triangle ACD. Also, since E and F are the circumcenters of triangles ACD and BCD, we have CE = AE and BF = CF, so that the triangles CEA and BFC are isosceles. Now, by the sum of angles in triangle CDA, we have < CDA = 180° - < DCA - < CAD = 180° - < UCA - A = 180° - (90° - B) - A = 90° + (B - A). Consequently, < CDB = 180° - < CDA = 180° - (90° + (B - A)) = 90° - (B - A). Thus, < ECA = < CDA - 90° = (90° + (B - A)) - 90° = B - A; < EAC = < CDA - 90° = (90° + (B - A)) - 90° = B - A; < FBC = 90° - < CDB = 90° - (90° - (B - A)) = B - A; < FCB = 90° - < CDB = 90° - (90° - (B - A)) = B - A. Hence, in particular, < ECA = < FCB and < EAC = < FBC. Thus, the triangles ECA and FCB are similar. This yields CE : CA = CF : CB. On the other hand, the equality < ECA = < FCB yields < ECF = < ECB - < FCB = < ECB - < ECA = < ACB. This, combined with CE : CA = CF : CB, shows that the triangles CEF and CAB are similar, so that EF : AB = CE : CA. On the other hand, the equation AK = a + b implies CK = AK - CA = (a + b) - b = a = CB, and similarly CL = CA. This, together with < KCL = < BCA, shows that the triangles CLK and CAB are congruent. Consequently, KL = AB. Hence, the equation EF : AB = CE : CA becomes EF : KL = CE : CA. The problem asks us to prove that the quadrilateral EFKL is a rectangle if and only if B - A = 60°. Hence, in order to solve the problem, it is enough to show the following two assertions: Assertion 1. If B - A = 60°, then the quadrilateral EFKL is a rectangle. Assertion 2. If the quadrilateral EFKL is a rectangle, then B - A = 60°. Proof of Assertion 2. If the quadrilateral EFKL is a rectangle, then EF = KL. Since EF : KL = CE : CA, this yields CE = CA. Hence, the triangle CEA, which is already isosceles with CE = AE, must be equilateral. Hence, < ECA = 60°. Since < ECA = B - A, we therefore obtain B - A = 60°. This proves Assertion 2. Proof of Assertion 1. If B - A = 60°, then, since < ECA = B - A, we have < ECA = 60°. Hence, the triangle CEA, which is already isosceles with CE = AE, must be equilateral. Thus, CE = CA. Since EF : KL = CE : CA, this yields EF = KL. On the other hand, we know that the triangles CEF and CAB are similar, and that the triangles CLK and CAB are congruent. Thus, the triangles CEF and CLK are similar. Moreover, these two triangles must be congruent, since EF = KL. Hence, CE = CL and CF = CK. Thus, the triangles ECL and FCK are isosceles. The base angle of the isosceles triangle ECL equals $\measuredangle CEL=\frac{180^{\circ}-\measuredangle ECL}{2}=\frac{\measuredangle BCE}{2}=\frac{\measuredangle BCA+\measuredangle ECA}{2}$ $=\frac{C+\left(B-A\right)}{2}$ (since < BCA = C and < ECA = B - A) $=\frac{\left(B+C\right)-A}{2}=\frac{\left(180^{\circ}-A\right)-A}{2}$ (since B + C = 180° - A by the sum of the angles in triangle ABC) $=\frac{180^{\circ}-2A}{2}=90^{\circ}-A$. Also, since the triangles CEF and CAB are similar, we have < CEF = < CAB. Thus, < FEL = < CEF + < CEL = < CAB + (90° - A) = A + (90° - A) = 90°. Similarly, < EFK = 90°. Since the triangles CEF and CLK are similar, we have < CEF = < CLK, and since the triangle ECL is isosceles with CE = CL, we have < CEL = < CLE. Thus, < FEL = < CEF + < CEL = < CLK + < CLE = < ELK. Consequently, < FEL = 90° implies < ELK = 90°. Similarly, < FKL = 90°. Altogether, we have obtained < FEL = 90°, < EFK = 90°, < ELK = 90° and < FKL = 90°. Thus, the quadrilateral EFKL has four right angles, so that it must be a rectangle. This proves Assertion 1. Now, as both Assertions 1 and 2 are proven, the problem is solved. Darij

Incidentally, this was also Taiwan 2nd TST 2005 final exam problem 5.

Bye... Sayantan....

In case you did not notice that $\triangle AFC$ was equilateral - see Grobber's, as I did, then from $\triangle ABO\sim\triangle ADF$ get $\frac{DF}{BO}=\frac{AD}{AB}$, and, with $DF=AC, BO-AO$ getting $AO\cdot AD=AB\cdot AC$. If the internal angle bisector of $\angle BAC$ intersects $BC$ and the circumcircle of $\Delta ABC$ (second time) at $M,N$ respectively, then we know that $AB\cdot AC=AM\cdot AN$, or $DOMN$ is cyclic; from $\angle DMN=\angle DON$ we get the same $\hat C-\hat B=60^\circ$. Best regards, sunken rock

Another beautiful problem from Hojo Lee.

Solution from Twitch solves ISL stream: We start with a few observations which are always true regardless of the condition. Quadrilateral $HGCB$ is always an isosceles trapezoid, and in particular $BC = GH$. By angle chasing $\overline{AO} \perp \overline{GH}$ always holds. (One clean way to see this is to note that $\overline{GH}$ and $\overline{BC}$ are antiparallel through $\angle A$.) This implies $\overline{EF} \parallel \overline{GH}$. By Salmon theorem, we always have \[ \triangle AEF \overset{+}{\sim} \triangle ABC. \] [asy][asy] size(6cm); pair B = dir(200); pair C = dir(-20); pair A = dir(30); pair O = origin; pair D = extension(A, O, B, C); pair E = circumcenter(A, B, D); pair F = circumcenter(A, C, D); filldraw(A--B--C--cycle, invisible, blue); draw(A--D, blue); filldraw(A--E--F--cycle, invisible, red); pair G = A+(A-B)*abs(C-A)/abs(B-A); pair H = A+(A-C)*abs(B-A)/abs(C-A); filldraw(A--C--G--cycle, invisible, blue); filldraw(A--B--H--cycle, invisible, blue); filldraw(A--G--H--cycle, invisible, blue); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A$", A, dir(330)); dot("$O$", O, dir(270)); dot("$D$", D, dir(270)); dot("$E$", E, dir(E)); dot("$F$", F, dir(270)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); /* TSQ Source: !size(10cm); B = dir 200 C = dir -20 A = dir 30 R330 O = origin R270 D = extension A O B C R270 E = circumcenter A B D F = circumcenter A C D R270 A--B--C--cycle 0.1 lightblue / blue A--D blue A--E--F--cycle 0.1 lightred / red G = A+(A-B)*abs(C-A)/abs(B-A) H = A+(A-C)*abs(B-A)/abs(C-A) A--C--G--cycle 0.1 lightcyan / blue A--B--H--cycle 0.1 lightcyan / blue A--G--H--cycle 0.1 lightblue / blue */ [/asy][/asy] We begin now with: Claim: We have $\triangle AEF \cong \triangle ABC$ if and only if $\angle C - \angle B = 60^{\circ}$. Proof. The congruence just means $FA = AC$. Since $FA = FC$ always, triangle $AFC$ is equilateral if and only if $\angle AFC = 60^{\circ} \iff \angle ADC = 30^{\circ}$. As $\angle ADC = (90^{\circ} - \angle C) + \angle B$, and the result follows. $\blacksquare$ Claim: We have $EFGH$ is a parallelogram if and only if $\triangle AEF \cong \triangle ABC$. Proof. Since we already know $\overline{EF} \parallel \overline{GH}$, the parallelogram condition is equivalent to $EF = GH$, but as $GH = BC$ we get the earlier congruence. $\blacksquare$ It remains only to show that if $EFGH$ is a parallelogram then it is also a rectangle. In the situation of the claims, note that $EG = FH$ by symmetry through the oppositely congruent triangles $\triangle AEF$ and $\triangle AHG$ as needed.

@v_Enhance what is the name of the twitch channel?

weaving2 wrote: @v_Enhance what is the name of the twitch channel? https://twitch.tv/vEnhance which normally runs Friday 8pm ET. You can see a schedule at https://web.evanchen.cc/videos.html.

Quote: By Salmon theorem, we always have\[ \triangle AEF \overset{+}{\sim} \triangle ABC. \] Why is this the case? Can you provide any link to further material on the subject?

$EFGH$ is a rectangle if and only if $EFGH$ is a rectangle.

Assume GHEF is rectangle we will prove ∠C - ∠B = 60. EF = GH = BC and we have AEF and ABC are similar so ABC and AEF are congruent so AEB and AFC are regular triangles. ∠AFC = 60 so ∠ADC = 30. ∠ADC = ∠B + ∠OAB = ∠B + 90 - ∠C so ∠C - ∠B = 90 - 30 = 60 as wanted. Assume ∠C - ∠B = 60 we will prove GHEF is rectangle. ∠ADC = ∠B + 90 - ∠C = 30 so ∠AFC = 60 and FA = FC so AFC is regular triangle. we have AEF and ABC are similar and AC = AF so AEF and ABC are congruent so EF = BC = GH. ∠BAF = 180 - 2C so ∠FGA = 90 - ∠C. ∠HGF = ∠HGA + ∠FGA = ∠C + 90 - ∠C = 90. same way ∠GHE = 90 so HE || GF so HEFG is rectangle.

Mahdi_Mashayekhi wrote: we have AEF and ABC are similar how do you get that they are similar?

Inefficient angle chasing solution... will improve later if I have time Let $\angle BAC=\alpha$ and so on. Notice $\overline{AO}\perp\overline{EF}$ since $\overline{AO}$ is the radical axis of $(ABD)$ and $(ACD).$ Also, $\overline{AO}\perp\overline{GH}$ as $$\measuredangle(\overline{AO},\overline{GH})=\measuredangle(\overline{AO},\overline{AG})+\measuredangle AGH=\measuredangle BCA+\measuredangle OAB=90.$$Finally, $$\angle CDA=\measuredangle BCA+\measuredangle OCA=\measuredangle BCA+90-\measuredangle ABC=90-(\measuredangle ACB-\measuredangle CBA).$$ If Direction: $\gamma-\beta=60$ implies $EFGH$ is a rectangle. Proof. Notice $\triangle AEB$ is equilateral as $$\angle BEA=2\angle CDA=2(90-60)=60.$$Hence, $$\angle EAO=60+90-\angle ACB=\angle ACB-\angle CBA+90-\angle ACB=90-\angle CBA=\angle OAC$$so $\angle AEH=\tfrac{1}{2}\angle EAC=\angle OAC$ and $\overline{EH}\parallel\overline{AO}.$ Similarly, $\overline{FG}\parallel\overline{AO}.$ $\blacksquare$ Only If Direction: $EFGH$ is a rectangle implies $\gamma-\beta=60.$ Proof. We know $\triangle AEF\cong\triangle DEF$ so $$\angle AFE=\tfrac{1}{2}\angle AFD=\angle ACB.$$Similarly, $\angle FEA=\angle CBA$ so $\triangle AEF\sim\triangle ABC.$ Then, $\triangle AEF\cong\triangle AHG$ so $\triangle AEB$ is equilateral. Thus, $$90-(\measuredangle ACB-\measuredangle CBA)=\angle CDA=\tfrac{1}{2}\cdot 60=30.$$$\blacksquare$ $\square$

This took me way too long It is easy to see that $BCGH$ is a isosceles trapezoid. Let $X$ be $OA\cap GH.$ We have \[\angle GAX=\angle OAB=90^\circ-\angle ACB=90^\circ-\angle AGX.\]Thus, $AX\perp GH.$ We also have $EF\perp AD$ so $EF\parallel GH.$ It is clear why F is inside and E is outside so now, $\angle CFD=2\angle CAD=\angle COD$ so $CFOD$ is cyclic. Similarly, $BEOD$ is cyclic. We have \[\angle FOE+\angle FAE=360^\circ-\angle FOD-\angle DOE+\angle FDE\]\[=\angle FCD+\angle EBD+180^\circ-\angle FDC-\angle EDB=180^\circ\]which implies that $AFOE$ is cyclic. We have $\angle ACB-\angle ABC= 90^\circ-\angle OAB-(90^\circ-\angle OAC)=\angle OAC-\angle OAB.$ What's important to see is that \[\angle OAE=\angle OFE=\angle FOD-90^\circ=90^\circ-\angle FCD=\angle OAC\]and similarly, $\angle OAF=\angle OAB.$ Thus, $\angle OAC-\angle OAB=\angle OAC-\angle OAF=\angle CAF.$ Therefore, we have \[\angle ACB=\angle ABC\iff \triangle ACF \text{ is equilateral}\]We know that $\triangle ABC\sim \triangle AEF$ and $\triangle ABC\cong \triangle AHG$ so $\triangle ACF$ is equilateral $\iff$ $\triangle AFE\cong \triangle AHG.$ Since $EF\parallel GH$, $\triangle AFE\cong\triangle AHG\iff EFGH$ rectangle, as desired. Numbertheorydog wrote: Mahdi_Mashayekhi wrote: we have AEF and ABC are similar how do you get that they are similar? angle chasing.

Solved with proxima1681. Solution: Let $D' = AO \cap GH$. As $AD$ is the radical axis of $\odot(ADC)$ and $\odot(ADB)$, we get $EF \perp AO$. It is also easy see that $\triangle ABC \cong \triangle AHG$. [asy][asy] import graph; size(11cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(14); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7, xmax = 6, ymin = -4.184372376555464, ymax = 10.16202343462137; /* image dimensions */ pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen ffqqff = rgb(1.,0.,1.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((-0.8535898390715775,2.4107656245533)--(4.846554211506446,-2.620429350613437)--(-3.14,-2.7)--cycle, linewidth(1.1) + yqqqyq); draw((-0.8535898390715775,2.4107656245533)--(-5.051246248590594,6.115799681176306)--(2.251206267261163,9.350852527005179)--cycle, linewidth(1.1) + yqqqyq); draw((-0.8535898390715775,2.4107656245533)--(-0.895059345715962,-0.6375021281980235)--(3.0810645932710234,1.12395692083598)--cycle, linewidth(1.1) + ffqqff); draw((-2.6885720014591716,6.552852553843345)--(-2.1789546962989768,6.77861765259111)--(-2.4047197950467414,7.288234957751304)--(-2.9143371002069363,7.0624698590035395)--cycle, linewidth(1.1) + qqwuqq); draw((0.7780870069641727,0.10371691540088587)--(0.5523219082164079,0.6133342205610806)--(0.04270460305621318,0.3875691218133157)--(0.268469701803978,-0.12204818334687892)--cycle, linewidth(1.1) + qqwuqq); /* draw figures */ draw((-0.8535898390715775,2.4107656245533)--(4.846554211506446,-2.620429350613437), linewidth(1.1) + yqqqyq); draw((4.846554211506446,-2.620429350613437)--(-3.14,-2.7), linewidth(1.1) + yqqqyq); draw((-3.14,-2.7)--(-0.8535898390715775,2.4107656245533), linewidth(1.1) + yqqqyq); draw((-0.8535898390715775,2.4107656245533)--(1.3905292426795337,-2.654861991247058), linewidth(1.1) + blue); draw((-0.8535898390715775,2.4107656245533)--(-5.051246248590594,6.115799681176306), linewidth(1.1) + yqqqyq); draw((-5.051246248590594,6.115799681176306)--(2.251206267261163,9.350852527005179), linewidth(1.1) + yqqqyq); draw((2.251206267261163,9.350852527005179)--(-0.8535898390715775,2.4107656245533), linewidth(1.1) + yqqqyq); draw((-2.9143371002069363,7.0624698590035395)--(-0.8535898390715775,2.4107656245533), linewidth(1.1) + blue); draw((-0.8535898390715775,2.4107656245533)--(-0.895059345715962,-0.6375021281980235), linewidth(1.1) + ffqqff); draw((-0.895059345715962,-0.6375021281980235)--(3.0810645932710234,1.12395692083598), linewidth(1.1) + ffqqff); draw((3.0810645932710234,1.12395692083598)--(-0.8535898390715775,2.4107656245533), linewidth(1.1) + ffqqff); draw((-0.895059345715962,-0.6375021281980235)--(-3.14,-2.7), linewidth(1.1) + ffqqff); draw((-5.051246248590594,6.115799681176306)--(-0.895059345715962,-0.6375021281980235), linewidth(1.1) + blue); draw((3.0810645932710234,1.12395692083598)--(2.251206267261163,9.350852527005179), linewidth(1.1) + blue); /* dots and labels */ dot((-0.8535898390715775,2.4107656245533),dotstyle); label("$A$", (-1.6,2.1217356723134735), NE * labelscalefactor); dot((4.846554211506446,-2.620429350613437),dotstyle); label("$B$", (5.163488257557553,-2.712947165152712), NE * labelscalefactor); dot((-3.14,-2.7),dotstyle); label("$C$", (-3.796440261877091,-2.844324416170815), NE * labelscalefactor); dot((-5.051246248590594,6.115799681176306),dotstyle); label("$G$", (-5.819649927555882,6.06305320285656), NE * labelscalefactor); dot((2.251206267261163,9.350852527005179),dotstyle); label("$H$", (2.3520150857701427,9.610238980345336), NE * labelscalefactor); dot((0.8408620061861096,-1.4141036218163907),linewidth(4.pt) + dotstyle); label("$O$", (1.1170689261999718,-1.5305519059897863), NE * labelscalefactor); dot((1.3905292426795337,-2.654861991247058),linewidth(4.pt) + dotstyle); label("$D$", (1.3798234282361783,-3.2384561692251235), NE * labelscalefactor); dot((-0.895059345715962,-0.6375021281980235),linewidth(4.pt) + dotstyle); label("$F$", (-1.0900688909041634,-1.2677974039535806), NE * labelscalefactor); dot((3.0810645932710234,1.12395692083598),linewidth(4.pt) + dotstyle); label("$E$", (3.1928294922860037,1.3334721662048563), NE * labelscalefactor); dot((-2.9143371002069363,7.0624698590035395),linewidth(4.pt) + dotstyle); label("$D'$", (-3.139554006786575,7.481927513852071), NE * labelscalefactor); dot((0.268469701803978,-0.12204818334687892),linewidth(4.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Claim: $\triangle AEF \sim \triangle ABC$ and $EF \parallel HG$. Proof: Observe \[\angle AFE = \frac{1}{2} \angle AFD = \angle ACB.\]$\angle AEF = \angle ABC$ follows symmetrically proving the similarity. For the parallel part, \begin{align*} \angle AD'G & = \angle D'HA + \angle D'AH \\ & = \angle ABC + \angle CAD \\ & = 90^\circ = \angle(FE,AD) \end{align*}and the claim is proven. $\square$ We first prove the if direction. If $EFGH$ is a rectangle, then \[\overline{EF} = \overline{GH} = \overline{BC}.\]This would give $\triangle ABC \cong \triangle AEF$. This would give $\overline{AF} = \overline{FC} = \overline{AC}$ giving $\triangle AFC$ equilateral. It is not hard to compute $\angle AFC = 2(90^\circ -C +B)$. Setting this equal to $60^\circ$, we would get $C-B = 60^\circ$ as desired. For the iff direction, assume $C-B = 60^\circ$. Analogous calculations as above, again reveal that $\triangle AFC$ and $\triangle AEB$ as equilateral. Note that $\triangle AEH$ and $\triangle AFG$ are isosceles. These triangles would also imply $\triangle AEF \cong \triangle ACB$. With some angle chasing one can compute \[\angle AFG = \frac{1}{2} A - 30^\circ \qquad \text{and} \qquad \angle AEH = \frac{1}{2} A + 30^\circ.\]Finally observe that \[\angle GFE = C + \frac{1}{2} A - 30^\circ = 90 ^\circ = \angle HEF\]which proves that $EFGH$ is indeed a rectangle and we are done. $\blacksquare$

First we prove the necessity: It's clear that $$30=C-60+90-C=ABC+OAB=ADC\implies AFC=60\implies AG=AC=FA.$$Now note the cyclicislscelestrapezoid since $$HAB=GAC,BHA=HBA=90-HAB/2=90-GAC/2=AGC=ACG,$$and that $AFD=2C\implies DAF=90-C=BAO$. Now, $$GFE=GFA+EFA=90-DAF+AGF=90-1/2BAF+90-GAF=90,$$$HGF=HGA+AGF=C+90-C=90,$ where the last step follows from knowing that $$AGF=(180-GAF)/2=FAB/2=FAD=90-C.$$Finally, we note that $AEF\cong ABC$, which follows immediately from $$FA=FC, EAF=EAB+BAF=60+BAF=FAC+BAF=BAC,AFE=90-AFG=90-AGF=C.$$It is now evident that EF=BC=HG ($BAC\cong HAG$), whence the already known right trapezoid (EFG and HGF are 90 degrees) has EF=HG which makes it a rectangle! $\blacksquare$ As for sufficiency, if EFGH is a rectangle, remark that BC=HG=EF, and $$AFE=AFD/2=ACB,AEF=AED/2=ABC\stackrel{SAS}{\implies}ABC\cong AEF\implies AF=AC,$$whence AFC is equilateral, and $$60=AFC=2ADC=2(B+BAO)=2(B+90-C)=180+2B-2C\implies C-B=60.$$$\blacksquare$ I'm really happy about this solution because it only took half an hour and it was straightforward, a very nice problem, and I did it on my own, with some nice observations! Obviously my necessity was overkill but I can't be bothered to shorten it lol

how do you come up with problems like these

Solved with GrantStar First, let $P$ be the intersection of the altitude from $A$ to $BC$ with the circumcircle of $ABC.$ Notice that since $EF$ and $OF$ are the perpendicular bisectors of $AD$ and $AC,$ we have $\measuredangle AEF=\frac{1}{2} \measuredangle AED=\measuredangle ABC=\frac{1}{2} \measuredangle AOC=\measuredangle AOF,$ so $AEOF$ is cyclic and $AO \perp EF.$ Also, by symmetry, we have $\measuredangle AFE=\measuredangle ACB,$ so $\triangle AEF$ and $\triangle ABC$ are similar and similarly oriented. Thus, there exists a spiral similarity at $A$ sending $\triangle AEF$ to $\triangle ABC.$ Notice that this spiral similarity also sends $O$ to $P,$ so we have $\frac{EF}{BC}=\frac{AO}{AP}.$ Next, Reim on lines $AB,AC$ and quadrilaterals $AABC$ and $BCGH$ gives us that $GH$ is parallel to the tangent to the circumcircle of $ABC$ at $A,$ so $GH \perp AD$ and $GH \parallel EF.$ Also, we have $\angle EAD=90-\frac{1}{2} \angle AED=90-\angle ABC=90-\frac{1}{2} \angle AOC=\angle OAC,$ so $AD$ bisects $\angle EAC,$ so the tangent to the circumcircle of $ABC$ at $A$ bisects $\angle EAH,$ and thus also bisects $\angle GAF$ by symmetry. Therefore, $EFGH$ being a rectangle is equivalent to $EF=GH.$ However, we know $GH=BC,$ and $BC=EF$ if and only if $AO=AP.$ Since $OA=OP,$ this holds if and only if $AOP$ is equilateral, or equivalently $\angle OAP=60.$ Now, let $A'$ be the point such that $AA'BC$ is an isosceles trapezoid with $AA' \parallel BC,$ and let $AO$ intersect the circumcircle of $ABC$ again at $Q.$ Then we have $\angle OAP=\angle QAP=\angle A'CA=\angle ACB-\angle BCA'=\angle ACB-\angle ABC,$ so we are done.

Obviously $EF$ is the perpendicular bisector of $AD$ and $OE\perp AB$, $OF\perp AC$. For convenience let $L$ be the foot of $A$ to $BC$. Now \[\measuredangle DAE=\measuredangle LAB\]\[\measuredangle DAF=\measuredangle LAC\]so $\triangle AEF\sim \triangle ABC$. Now we prove that $EF\parallel GH$. Consider the acute angles formed by each of the lines with $BC$. It suffices to show: \[90^{\circ}-\angle ADC=180^{\circ}-(\angle A+2\angle B)\]\[90^{\circ}-(\angle B+90^{\circ}-\angle C)=180^{\circ}-(\angle A+2\angle B)\]which is true. Now since $\triangle AEF\sim \triangle ABC$, $\triangle ABC\cong \triangle AHG$ we must have $AF=AG=AC$ in order to have $EF=GH$. So $\angle AFC=60^{\circ}$, $\angle ADC=30^{\circ}$, so \[\angle ADC=\angle B+90^{\circ}-\angle C=30^{\circ}\implies \angle C-\angle B=60^{\circ}\]and we are done.

Since $\triangle GAB \cong \triangle CAH$, we have $GH \parallel BC$ due to $\angle ABG = \angle AHC$. If $EFGH$ is a rectangle, then $GH = EF$. Then consider the homothety $\mathcal{H}$ sending $(ABD) \to (ACD)$. Then $\mathcal{H}(E) = F$, and $\mathcal{H}(B) = C$, so $\triangle{AEF} \cong \triangle{ABC}$. Then notice that if $\triangle{AEF} \cong \triangle{ABC}$, then $AF = AC$ which implies that $\triangle AFC$ is equilateral. From here, we find $\angle FAC = \angle FCA = 60^{\circ} \implies \angle ADC = 30^{\circ}$. We can angle chase to find $\angle B = 90^{\circ} = \angle C + 30^{\circ}$, so we are done.

Headsolved! Typed up without a diagram either, so there might be some mistakes in point names lol. First, we will prove that $HG$ and $EF$ are always parallel. Note that $HG$ and $BC$ are reflections over the $A$-external angle bisector, and lines $AO$ and $AH$ are reflections over the $A$-external angle bisector (since they are isogonal). Since $EF\perp AO$, we want to show that $AH\perp BC$, which is obvious. Now suppose $EF=GH$. Then note that the projections of $E$ and $F$ onto $BC$ have distance exactly half of $BC$. Thus the angle bietween $EF$ and $BC$ is $60^{\circ}$, so $\angle ADC=30^{\circ}$, so \[180^{\circ}=30^{\circ}+\angle DAC+\angle C=30^{\circ}+(90^{\circ}-\angle B)+\angle C\Longrightarrow \angle C-\angle B=60^{\circ}.\] Conversely, if $\angle C-\angle B=60^{\circ}$, the previous paragraph is all reversible, so we still get $EF=GH$, so $EFGH$ is a parallelogram. Also, $\angle AFC=60^{\circ}$, so $AF=AC=AG$. Since $\angle BAD=\angle DAF=90^{\circ}-\angle C$, so $\angle FAG=2\angle C$. But note that since $AF=AG$, we get $\angle AFG=90^{\circ}-\angle C=\angle DAF$, so $AO\parallel FG$, so $EF\perp FG$, so we get that $EFGH$ is a rectangle. $\blacksquare$