I cant upload the solutoon as the cgaracters have brackets, aops says invalid chars But it was again usage of isogonality of circum and orthocenter

One liner. Apply isgonal line lemma for the pair of lines $(AH,AD)$ w.r.t $\angle B_1AC_1$ where $H=BB_1\cap CC_1$. Done $\blacksquare$.

Is there any Angle Chasing Solution? Sharygin 2020 CR P5 wrote: Let $BB_1$, $CC_1$ be the altitudes of triangle $ABC$, and $AD$ be the diameter of its circumcircle. The lines $BB_1$ and $DC_1$ meet at point $E$, the lines $CC_1$ and $DB_1$ meet at point $F$. Prove that $\angle CAE = \angle BAF$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -11.45, xmax = 11.45, ymin = -6.11, ymax = 6.11; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-1.97,2.91)--(-3.95,-2.29), linewidth(0.4) + blue); draw((-3.95,-2.29)--(2.73,-2.29), linewidth(0.4) + wrwrwr); draw((2.73,-2.29)--(-1.97,2.91), linewidth(0.4) + blue); draw(circle((-0.61,-0.5848076923076923), 3.7501041060499927), linewidth(0.4) + red); draw((-3.95,-2.29)--(-0.2267089234835029,0.9812524259817477), linewidth(0.4) + wrwrwr); draw((2.73,-2.29)--(-3.107190877254641,-0.07656189986067341), linewidth(0.4) + wrwrwr); draw((0.75,-4.079615384615384)--(-0.2267089234835029,0.9812524259817477), linewidth(0.4) + wrwrwr); draw((0.75,-4.079615384615384)--(-3.107190877254641,-0.07656189986067341), linewidth(0.4) + wrwrwr); draw((-1.97,2.91)--(0.75,-4.079615384615384), linewidth(0.4) + wrwrwr); draw((-1.97,2.91)--(-2.338589668237001,-0.8742283483343429), linewidth(0.4) + blue); draw((0.2210048493121505,-1.3385996472657353)--(-1.97,2.91), linewidth(0.4) + blue); /* dots and labels */ dot((-1.97,2.91),dotstyle); label("$A$", (-2.11,3.19), NE * labelscalefactor); dot((-3.95,-2.29),dotstyle); label("$B$", (-4.45,-2.57), NE * labelscalefactor); dot((2.73,-2.29),dotstyle); label("$C$", (2.81,-2.09), NE * labelscalefactor); dot((-1.89,-0.51),dotstyle); label("$H$", (-1.89,-0.97), NE * labelscalefactor); dot((-0.2267089234835029,0.9812524259817477),dotstyle); label("$B_1$", (-0.15,1.19), NE * labelscalefactor); dot((-3.107190877254641,-0.07656189986067341),dotstyle); label("$C_1$", (-3.63,0.05), NE * labelscalefactor); dot((-0.61,-0.5848076923076923),linewidth(4pt) + dotstyle); label("$O$", (-1.01,-0.55), NE * labelscalefactor); dot((0.75,-4.079615384615384),dotstyle); label("$D$", (0.85,-4.49), NE * labelscalefactor); dot((-2.338589668237001,-0.8742283483343429),linewidth(4pt) + dotstyle); label("$E$", (-2.43,-1.39), NE * labelscalefactor); dot((0.2210048493121505,-1.3385996472657353),linewidth(4pt) + dotstyle); label("$F$", (0.45,-1.93), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Isogonal Line Lemma:-Consider two arbitrary points $P,Q$ in the plain of $\triangle ABC$. Let $P',Q'$ be the isogonal conjugate of $P,Q$ WRT $\triangle ABC$ respectively. Let $PQ\cap P'Q'=S, P'Q\cap PQ'=T$ prove that $T$ is isogonal conjugate of $S$ WRT $\triangle ABC$. This problem actually tells us to prove that $AE$ and $AF$ are isogonal. Let $BB_1\cap CC_1=H$. So, $\angle CAD=90^\circ-\angle ADC=90^\circ-\angle B=\angle BAH$. So, $\{AH,AD\}$ and $\{AB_1,AC_1\}$ are isogonal pairs. and $E=HB_1\cap DC_1$ and $F=HC_1\cap DB_1$. So, by Isogonal Line Lemma we conclude that $\{AE,AF\}$ are isogonal WRT $\triangle ABC$. $\blacksquare$

We can easily generalize this. Attached

Attachments:
P5.pdf (77kb)

From where did y'all get to know this Lemma? And also I have heard a lot about Reim's theorem, does anybody have any handouts/suggested books to learn these two?

immoral solution: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -9.000853584705306, xmax = 7.662932447333153, ymin = -6.7817435165633295, ymax = 3.7038241955026066; /* image dimensions */ /* draw figures */ draw(circle((-2.29,-0.9114978455068345), 4.4917022251458425), linewidth(1)); draw((-6.51,-2.45)--(-2.4556994388912465,1.7667456285565915), linewidth(1)); draw((1.93,-2.45)--(-5.130733289458223,0.6706785124654949), linewidth(1)); draw((-2.4556994388912465,1.7667456285565915)--(-0.5869848657334074,-5.067833771757146), linewidth(1)); draw((-0.5869848657334074,-5.067833771757146)--(-5.130733289458223,0.6706785124654949), linewidth(1)); draw((-3.9930151342665923,0.16783377175714626)--(-0.5869848657334074,-5.067833771757146), linewidth(1)); draw((-6.51,-2.45)--(-0.5869848657334074,-5.067833771757146), linewidth(1)); draw((-0.5869848657334074,-5.067833771757146)--(1.93,-2.45), linewidth(1)); draw(circle((-2.29,-2.45), 4.22), linewidth(1)); draw((-3.993015134266591,3.2448380807434796)--(-6.51,-2.45), linewidth(1)); draw((-6.51,-2.45)--(1.93,-2.45), linewidth(1)); draw((1.93,-2.45)--(-3.993015134266591,3.2448380807434796), linewidth(1)); draw((-3.993015134266591,3.2448380807434796)--(-4.398584819551673,-0.25398601484329136), linewidth(1)); draw((-3.993015134266591,3.2448380807434796)--(-1.7471174775414104,-0.8248003175382232), linewidth(1)); draw((-2.4556994388912465,1.7667456285565915)--(-5.130733289458223,0.6706785124654949), linewidth(1)); /* dots and labels */ dot((1.93,-2.45),dotstyle); label("$C$", (2.093834243414295,-2.550534138976557), NE * labelscalefactor); dot((-6.51,-2.45),dotstyle); label("$B$", (-6.792793164010914,-2.572288428835615), NE * labelscalefactor); dot((-3.993015134266591,3.2448380807434796),dotstyle); label("$A$", (-3.95385833740384,3.3557555577576794), NE * labelscalefactor); dot((-2.4556994388912465,1.7667456285565915),linewidth(4pt) + dotstyle); label("$B_{1}$", (-2.4745666269878934,1.930849571989383), NE * labelscalefactor); dot((-5.130733289458223,0.6706785124654949),linewidth(4pt) + dotstyle); label("$C_{1}$", (-5.476658627537903,0.810503644247898), NE * labelscalefactor); dot((-3.9930151342665923,0.16783377175714626),linewidth(4pt) + dotstyle); label("$H$", (-4.095261221487717,0.31015497748956494), NE * labelscalefactor); dot((-0.5869848657334074,-5.067833771757146),linewidth(4pt) + dotstyle); label("$D$", (-0.5384348295317277,-4.976137458261519), NE * labelscalefactor); dot((-4.398584819551673,-0.25398601484329136),linewidth(4pt) + dotstyle); label("$E$", (-4.4650841490917035,-0.6144023415203984), NE * labelscalefactor); dot((-1.7471174775414104,-0.8248003175382232),linewidth(4pt) + dotstyle); label("$F$", (-1.7022893369913328,-0.7340509357452172), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Observe that by applying the Sine Law to triangles $\triangle HDB_1$ and $\triangle HDC_1,$ we find that $$\frac{\sin \angle B_1DH\sin \angle DC_1H}{\sin \angle HDC_1 \sin \angle HB_1D}=\frac{\frac{\sin\angle DHB_1}{B_1D\cdot HB_1}\cdot \frac{\sin\angle C_1DH}{C_1D\cdot HD}}{\frac{\sin\angle C_1HD}{C_1D\cdot HC_1}\cdot \frac{\sin\angle DHB_1}{B_1D\cdot HD}}=\frac{HC_1}{HB_1}=\frac{HB}{HC}=\frac{DC}{DB},$$where the last two equalities come from the well-known facts that $BCB_1C_1$ is a cyclic quadrilateral and $BHCD$ is a parallelogram. Next, we use the Sine Law again on $\triangle C_1B_1E,\triangle B_1C_1F,\triangle C_1BE,$ and $\triangle B_1CF,$ which gives $$\frac{\sin\angle C_1B_1H}{\sin\angle HC_1B_1}=\frac{\frac{\sin\angle DC_1B_1}{EB_1\cdot C_1E}}{\frac{\sin\angle C_1B_1D}{FC_1\cdot B_1F}}=\frac{FC_1\cdot \frac{FC\sin \angle ACC_1}{\sin \angle DB_1C}\cdot \sin\angle DC_1B_1}{EB_1\cdot \frac{EB\sin\angle B_1BA}{\sin\angle BC_1D}\cdot \sin\angle C_1B_1D}=\frac{FC_1\cdot FC\cdot DB}{EB_1\cdot EB\cdot DC}$$since $\dfrac{DB_1}{\sin\angle DC_1B_1}=\dfrac{DC_1}{\sin\angle C_1B_1D}.$ Finally, by Ceva's Theorem applied to triangle $\triangle DB_1C_1,$ we find that $$1 = \dfrac{\sin \angle B_1DH\sin \angle C_1B_1H\sin \angle DC_1H}{\sin \angle HDC_1 \sin \angle HB_1D \sin \angle HC_1B_1}\implies \frac{FC}{EB}=\frac{EB_1}{FC_1}.$$Thus, since $\triangle BAB_1\stackrel{-}{\sim}\triangle CAC_1,$ we get $\triangle EAB\stackrel{-}{\sim}\triangle FAC\implies \angle FAC=\angle BAE\implies \angle EAC=\angle BAF$ as desired. $\blacksquare$

Am I the only one who angle bashed it? Problem 5. Let $BB_1, CC_1$ be the altitudes of triangle $ABC$, and $AD$ be the diameter of its circumcircle. The lines $BB_1$ and $DC_1$ meet at point $E$, the lines $CC_1$ and $DB_1$ meet at point $F$. Prove that $\angle CAE = \angle BAF$. \newline [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -14.35, xmax = 14.35, ymin = -10.87, ymax = 10.87; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqccqq = rgb(0,0.8,0); draw((-0.57,5.85)--(-4.35,-2.65)--(6.35,-2.89)--cycle, linewidth(0.8) + zzttqq); /* draw figures *//* special point *//* special point *//* special point */ draw((-0.57,5.85)--(-4.35,-2.65), linewidth(0.8) + zzttqq); draw((-4.35,-2.65)--(6.35,-2.89), linewidth(0.8) + zzttqq); draw((6.35,-2.89)--(-0.57,5.85), linewidth(0.8) + zzttqq); draw(circle((1.0628789904329765,0.033354990136864535), 6.041494266169846), linewidth(0.8)); draw((-0.57,5.85)--(2.695757980865953,-5.783290019726271), linewidth(0.8)); draw((6.35,-2.89)--(-2.672424727057584,1.1223253492091365), linewidth(0.8)); draw((2.1101761269452983,2.464922059320533)--(-4.35,-2.65), linewidth(0.8)); draw((-1.496343067499764,-0.3905828406291039)--(-0.57,5.85), linewidth(0.8) + qqccqq); draw((-0.57,5.85)--(2.3645196718492744,-1.1176334540694425), linewidth(0.8) + qqccqq); draw((-0.57,5.85)--(-4.171300769930015,3.050475339803007), linewidth(0.8) + linetype("4 4") + qqccqq); draw((-0.57,5.85)--(1.857616609865691,6.022348652215792), linewidth(0.8) + linetype("4 4") + qqccqq); draw((2.1101761269452983,2.464922059320533)--(-4.171300769930015,3.050475339803007), linewidth(0.8) + blue); draw((-2.672424727057584,1.1223253492091365)--(1.857616609865691,6.022348652215792), linewidth(0.8) + blue); draw((-4.171300769930015,3.050475339803007)--(1.857616609865691,6.022348652215792), linewidth(0.8) + linetype("0 3 4 3") + red); draw((-2.672424727057584,1.1223253492091365)--(2.1101761269452983,2.464922059320533), linewidth(0.8) + linetype("0 3 4 3") + red); draw(circle((-0.9112047545879015,4.038097638424919), 3.406409228789864), linewidth(0.8) + linetype("4 4") + blue); draw((-4.171300769930015,3.050475339803007)--(2.695757980865953,-5.783290019726271), linewidth(0.8) + red); draw((2.695757980865953,-5.783290019726271)--(1.857616609865691,6.022348652215792), linewidth(0.8) + red); draw((-4.35,-2.65)--(2.695757980865953,-5.783290019726271), linewidth(0.8)); draw((2.695757980865953,-5.783290019726271)--(6.35,-2.89), linewidth(0.8)); /* dots and labels */ label("$A$", (-0.87,6.25), NE * labelscalefactor); label("$B$", (-4.81,-2.91), NE * labelscalefactor); label("$C$", (6.49,-2.91), NE * labelscalefactor); dot((1.0628789904329765,0.033354990136864535),linewidth(4pt) + dotstyle); label("$O$", (1.15,0.19), NE * labelscalefactor); dot((2.695757980865953,-5.783290019726271),linewidth(4pt) + dotstyle); label("$D$", (2.77,-5.63), NE * labelscalefactor); dot((-2.672424727057584,1.1223253492091365),linewidth(4pt) + dotstyle); label("$C_1$", (-3.15,1.15), NE * labelscalefactor); dot((2.1101761269452983,2.464922059320533),linewidth(4pt) + dotstyle); label("$B_1$", (2.11,2.73), NE * labelscalefactor); dot((-1.496343067499764,-0.3905828406291039),linewidth(4pt) + dotstyle); label("$E$", (-1.57,-0.77), NE * labelscalefactor); dot((2.3645196718492744,-1.1176334540694425),linewidth(4pt) + dotstyle); label("$F$", (2.45,-0.95), NE * labelscalefactor); dot((-4.171300769930015,3.050475339803007),linewidth(4pt) + dotstyle); label("$G$", (-4.55,3.01), NE * labelscalefactor); dot((1.857616609865691,6.022348652215792),linewidth(4pt) + dotstyle); label("$H$", (1.93,6.19), NE * labelscalefactor); dot((1.2464048496185782,-0.6204012154774389),linewidth(4pt) + dotstyle); label("$I$", (1.33,-0.47), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution. Notations: Let $DC_1$ and $DB_1$ meet the circumcircle of $\triangle ABC$ for the second time at $G$ and $H$ respectively. Also, assume $AD$ and $CC_1$ intersect at $I$. We will also denote, circumcircle of triangle $XYZ$ by $(XYZ)$. For sake of simplcity, we denote, $\angle ABC=b, \angle CAB =a, \angle ACB =c$. Proof: We begin by noting that $CC_1$ is parallel to $BD$ because, $\angle DBA=90=\angle CC_1A$ as $AD$ is diameter of $(ABC)$. Thus, $\angle CID=\angle IDB=\angle ADB=\angle ACB=c$. Now, let $\angle AHG=x\implies \angle GHD=90-x$ as $\angle AHD=90$. Thus, $\angle AHG=\angle ADG=x$, but $\angle CID= c\implies \angle IC_1D=c-x$. Also, we know that $\angle B_1BC=90-c$ and since, $BCB_1C_1$ is cyclic quadrilateral with diameter $BC$, we have, $\angle B_1BC=90-c=\angle B_1C_1C\implies \angle B_1C_1D=\angle B_1C_1C+\angle CC_1D=(90-c)+(c-x)=90-x=\angle GHB_1\implies GHB_1C_1$ is a cyclic quadrilateral. Thus, we get, $\angle C_1HB_1=\angle B_1GC_1$. Also, we have, $\angle AGC_1=\angle AGD=90=\angle AHD=\angle AHB_1\implies \angle AGC_1-\angle B_1GC_1=\angle AHB_1-C_1HB_1\implies \boxed{\angle AGB_1=\angle AHC_1}$. Now, we note that $AHFC_1$ and $AGEB_1$ are cyclic because, $\angle AHF=90=\angle AC_1F$ and $\angle AGE=\angle AB_1E$. Thus, we must have, $\angle AHC_1=\angle AFC_1$ and $\angle AGB_1=\angle AEB_1$. Nut, we also have $\angle AGB_1=\angle AHC_1$ and thus, we get $\angle AFC_1=\angle AEB_1\implies \boxed{\angle AFC=\angle AEB}$. Also, we have $\angle ABE=90-a=\angle ACF$. Thus, $\angle ABE +\angle AEB = \angle ACF +\angle AFC\implies \angle BAE=\angle CAF\implies \angle BAF =\angle CAE$ completing the proof. Feedback: A great problem for angle chasing .

Solution: Lemma : Let $AP,AS$ and $AQ,AR$ be two pairs of isogonal lines with respect to $\angle BAC$. Let $PR \cap QS = X$ and $PQ \cap RS = Y$. Then $AX,AY$ are isogonal line with respect to $\angle BAC$ Proof: Without loss of generality let $AY$ intersect $PR$ , $QS$ in $K,L$ respectively. Considering the perspectivity that sends line $PR \to QS$ ($Y$ is the centre of perspectivity). $\implies \frac{PK \cdot XR}{XK \cdot PR} = \frac { QL \cdot SX}{XL \cdot SQ}$ Now using the property of cross ratio w.r.t point $A$. Let $\angle PAQ = \angle RAS = x, \angle QAL = y, \angle XAR = z$ Using the cross ratios, $sin(x+y) \cdot sin z = siny \cdot sin (x+z)$ $\implies siny = sinz$ $\implies y=z$ (configuration matters here) $\implies \angle QAY = \angle RAX$ Let $H$ be the orthocentre of $\triangle ABC$ We see that AD passes through the circumcentre $O$ of $\triangle ABC$, and we know that $AH$ and $AO$ are isogonal with respect to $\angle BAC$ We know that $HB_1 \cap DC_1 = E $ and $HC_1 \cap DB_1 = F $ Applying this lemma on isogonal lines $AH$ , $AD$ and $AB$ , $AC$ , we get that $AE$ and $AF$ are isogonal with respect to $\angle BAC$ $\implies \angle CAE = \angle BAF $ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(26.18cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.3, xmax = 21.88, ymin = -10.46, ymax = 6.3; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pair A = (8.8,3.66), B = (6.88,-4.02), C = (18.6,-2.94), O = (12.544279475982533,-1.3560698689956334), D = (16.288558951965065,-6.372139737991266), H = (9.191441048034935,-0.5878602620087324), B_1 = (11.037432664756446,2.1531575931232103), C_1 = (7.823529411764706,-0.24588235294117497), F = (13.369048324273313,-1.6322620810683277); draw(A--B--C--cycle, linewidth(2.)); /* draw figures */ draw(A--B, linewidth(2.)); draw(B--C, linewidth(2.)); draw(C--A, linewidth(2.)); draw(circle(O, 6.25943972931363), linewidth(1.6) + green); draw(B--B_1, linewidth(2.)); draw(C_1--C, linewidth(2.)); draw(C_1--D, linewidth(2.)); draw(D--B_1, linewidth(2.)); /* dots and labels */ dot(A,ududff); label("$A$", (8.88,3.86), NE * labelscalefactor,ududff); dot(B,ududff); label("$B$", (6.96,-3.82), NE * labelscalefactor,ududff); dot(C,ududff); label("$C$", (18.68,-2.74), NE * labelscalefactor,ududff); dot(O,linewidth(3.pt) + uuuuuu); label("$O$", (12.62,-1.24), NE * labelscalefactor,uuuuuu); dot(D,linewidth(4.pt) + uuuuuu); label("$D$", (16.36,-6.22), NE * labelscalefactor,uuuuuu); dot(H,linewidth(4.pt) + uuuuuu); label("$H$", (9.28,-0.42), NE * labelscalefactor,uuuuuu); dot(B_1,linewidth(4.pt) + uuuuuu); label("$B_1$", (11.12,2.32), NE * labelscalefactor,uuuuuu); dot(C_1,linewidth(4.pt) + uuuuuu); label("$C_1$", (7.9,-0.08), NE * labelscalefactor,uuuuuu); dot((8.898038205214444,-1.0235190286209768),linewidth(4.pt) + uuuuuu); label("$E$", (8.98,-0.86), NE * labelscalefactor,uuuuuu); dot(F,linewidth(4.pt) + uuuuuu); label("$F$", (13.44,-1.48), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Geronimo_1501 wrote: From where did y'all get to know this Lemma? And also I have heard a lot about Reim's theorem, does anybody have any handouts/suggested books to learn these two? Bump..

Geronimo_1501 wrote: Geronimo_1501 wrote: From where did y'all get to know this Lemma? And also I have heard a lot about Reim's theorem, does anybody have any handouts/suggested books to learn these two? Bump.. You can see that Lemma Here, it can be generalized to Isotomic Conjugates too. Isogonal Line Lemma. Even Pg 90 of Geometry of Conic has description about this.

Geronimo_1501 wrote: Geronimo_1501 wrote: From where did y'all get to know this Lemma? And also I have heard a lot about Reim's theorem, does anybody have any handouts/suggested books to learn these two? Bump.. Lol reim's theorem is just a fashionable name, you need handouts on them?

You can also show this by the well-known lemma: Given cyclic quadrilateral $ABCD$ and with diagonals $AC$ and $BD$. Let $AC \cap BD=F$ and $AB \cap CD=E$. Then, $EF$ and the parallel from $E$ to the Newton-Gauss line of complete quadrilateral $EAFDBC$ are isogonal. In here $ABCD$ is $B_1CBC_1$.

LoL.No one has complex solution Every point is computable. ∆BAE ~ ∆CAF

Nymoldin wrote: LoL.No one has complex solution Every point is computable. ∆BAE ~ ∆CAF I had actually tried this. Isn't computing $E$ and $F$ like a lot of computation.

Geronimo_1501 wrote: From where did y'all get to know this Lemma? And also I have heard a lot about Reim's theorem, does anybody have any handouts/suggested books to learn these two? See attached a worksheet that I gave to my students in order to practice applying this lemma (sorry there are no solutions written in the worksheet). This lemma and other properties of isogonal lines will be added in the next version (v1.4) of my book A Beautiful Journey Through Olympiad Geometry, which will hopefully be released by the end of August.

Attachments:
2019-06-14 Advanced Geometry Worksheet 12.pdf (208kb)

Steff9 wrote: ... the next version (v1.4) of my book A Beautiful Journey Through Olympiad Geometry, which will hopefully be released by the end of August. I'm looking forward into it. It's a great book. I'm brooding on similar idea for years, but it will probably just remain as idea and you actually did it. Great respect. btw, that link to your site doesn't want to open. Maybe it just for me. I tried it before, always with the same result.

@above link works perfectly fine, maybe try other device @2above Your book is really nice

Pluto1708 wrote: We can easily generalize this. Attached We can also generalize using moving points. In the notation of Pluto1708's pdf, let $\ell=AQ$. Move $Q$ over $\ell$ and map to $Q_B$. Then, $Q\mapsto Q_B$ is a perspectivity. Let $Q_B'\in CP$ such that $AQ_C$ and $AQ_B'$ are isogonal. Then $Q\mapsto Q_C $ is also a perspectivity, $Q_C\mapsto AQ_C\mapsto AQ_B' \mapsto Q'B$ is also projective, because $AQ\mapsto AQ_B'$ is a reflection over the A-angle bisector. Therefore to show that $Q\mapsto Q_B$ and $Q\mapsto Q_B'$ coincide, it suffices to check three values of $Q$. But note that the cases $Q\equiv A$, $Q\equiv PB\cap \ell$ and $Q\equiv PC\cap \ell$ are trivial, so we are done. $\blacksquare$

zuss77 wrote: Steff9 wrote: ... the next version (v1.4) of my book A Beautiful Journey Through Olympiad Geometry, which will hopefully be released by the end of August. I'm looking forward into it. It's a great book. I'm brooding on similar idea for years, but it will probably just remain as idea and you actually did it. Great respect. @zuss77: v1.4 of the book is out, you can learn more about the Isogonal Lines Lemma in Chapter 24

Simple solution: Using only that $CD \parallel B_1H$ and $BD \parallel HF$ it follows that \[\frac{CF}{CH}=\frac{DF}{DB_1}=\frac{BH}{BB_1}\]and hence $CF=\frac{BH \cdot CH}{BB_1}$. Similarly, $BE=\frac{BH \cdot CH}{CC_1}$ and hence $\frac{BE}{CF}=\frac{BB_1}{CC_1}$. Since $ABB_1$ and $ACC_1$ are similar, then also $ABE$ and $ACF$ are similar and we are done.

ShamimAkhtar212 wrote: Geronimo_1501 wrote: Geronimo_1501 wrote: From where did y'all get to know this Lemma? And also I have heard a lot about Reim's theorem, does anybody have any handouts/suggested books to learn these two? Bump.. Lol reim's theorem is just a fashionable name, you need handouts on them? Hi! Can you please make a handout regarding the variations of Reims theorem?It would be nice because it is very useful and no handout are here based on it.