Lemma: $ CC_{1}\perp DJ$. Proof: Let $ w_{A}$ be the $ A$-excircle of $ \triangle ABC$, Let $ w_{C}$ be the circle diameter $ CJ$, let $ w$ be the circle diameter $ C_{1}J$. It is evident that: $ A_{1}$ and $ B_{1}$ lie on $ w_{C}$. Let $ w\cap w_{C}=J,E'$. Note that $ C_{1}C_{1}$, $ E'J$, $ A_{1}B_{1}$ are concurrent; they are radical axes of the three circles I mentioned earlier. But $ \angle CE'J=90$ from $ w$, and $ \angle CE'J=90$ from $ w_{C}$, so $ C_{1}E'C$ is a line; $ E$ and $ E'$ are equivalent since they are both the point that lies on $ CC_{1}$ and $ DJ$. End Lemma Let $ \angle A_{1}BD=x$ -$ \angle BA_{1}D=90-x$ since $ BD\perp DA_{1}$. -$ \angle CA_{1}B=\angle CBA_{1}=90-x$ by vertical angles and $ A_{1}C=A_{1}B$. -$ \angle JCA_{1}=x$, $ CA_{1}=CB_{1}$ and $ JA_{1}=JB_{1}$ so $ JC$ is the perpendicular bisector of $ A_{1}B_{1}$. -$ \angle A_{1}JC=\angle CEA_{1}=90-x$ since $ A_{1}C\perp A_{1}J$ and $ A_{1}EJC$ is cyclic. -$ \angle DEA_{1}=x$ Since $ DE\perp EC$. Hence $ DBEA_{1}$ is cyclic and $ \boxed{\angle BEA_{1}=90^\circ}$ Now another long angle chase. -$ \angle A_{1}JC=90-x=\angle CJB_{1}$ since $ A_{1}JC\cong B_{1}JC$ by SSS -$ \angle CJB_{1}=\angle CEB_{1}=90-x$ since $ CEJB_{1}$ is cyclic. -$ \angle DEB_{1}=90+x$ by addition. -$ \angle BCA=180-2x$ since $ \angle A_{1}CJ=\angle B_{1}CJ=x$ -$ \angle BAC=x$ by the sum of the angles in a triangle. Hence $ \angle DAB_{1}+\angle DEB_{1}=x+180-x=180$ so $ ADEB_{1}$ is cyclic. It follows that $ \angle ADB_{1}=\boxed{\angle AEB_{1}=90^\circ}$. Phew that was a lot of angle chasing.

it's obvious that $ AC=BC$. the polar of $ C$ pass through $ D$ so the polar of $ D$ pass through $ C$,polar of $ D$ also pass through $ C_{1}$ and is perpendicualr to $ I_{a}D$ hence $ C_{1}E$ pass through $ C$. $ \angle A_{1}I_{a}B_{1}=\angle ACB$ and $ \angle I_{a}EC=\angle CB_{1}I_{a}=90^{\circ}$ hence five points $ A_{1},C,B_{1},I_{a},E$ are on a circle . $ CI_{a}\parallel AB$ hence $ \angle CB_{1}E=\angle CI_{a}E=\angle I_{a}DC_{1}$ hence $ ADEB_{1}$ is cyclic also you can prove that $ DBEA_{1}$ is cyclic,too. hence $ \angle AEB_{1}=\angle BEA_{1}=90^{\circ}$

Lemma 1: $AC=BC$. It is known that $A_1 B_1$ is parallel to the bisector of $\angle{C}$. Since $A_1 B_1 \perp AB$, the bisector of $\angle{C}$ is perpendicular to $AB$. Hence $AC=BC$. Lemma 2: $BA_1 EJ$ is cyclic. Since $JC_1 \perp DC_1$ and $C_1 E \perp DJ$, $\triangle{DEC_1} \sim \triangle{DC_1 J}$. This yields that $DE \cdot DJ=DC_1^2$. By power of point $D$ with respect to the $A$-excircle, \[DE \cdot DJ=DC_1^2=DA_1 \cdot DB_1\] Hence $BA_1 EJ$ is cyclic by converse power of a point. Lemma 3: $A_1 EBD$ is cyclic. Note that since $B_1 J \perp AB_1$, \[\angle{A_1 ED}=\angle{JB_1 D}=90-\angle{AB_1 D}=\angle{BAC}=\angle{A_1 BD}\] Hence $A_1 EBD$ is cyclic. Lemma 4: $ADEB_1$ is cyclic. Note that \[\angle{B_1 ED}=\angle{A_1 ED}+\angle{B_1 EA_1}=\angle{CAB}+\angle{B_1 JA_1}=\angle{CAB} + 180-2\angle{CAB}=180-\angle{CAB}\] Hence $ADEB_1$ is cyclic. Therefore $\angle{AEB_{1}}=\angle{BEA_{1}}=90$.

Solution : Use radical lines.

[geogebra]4a3df5a7b7dfb5852c91d817df7006087e5e29e3[/geogebra] A solution by inversion!! Invert in the excircle Let $AJ\cap C_1B_1=A^'$ Consider the quadrilateral $AEA^'B_1$ Under inversion, $A \rightarrow A^'$ . . . ($\Delta JC_1B$ is right and $C_1E \bot DJ$) $E \rightarrow D$ . . . ($\Delta JC_1B$ is right and $C_1E \bot DJ$) $A^' \rightarrow A$ $B_1 \rightarrow B_1$ $ADA^'B_1$ is cyclic and as circles remain circles under inversion, $AEA^'B_1$ is cyclic Therefore, $\angle AEB_1= \angle AA^'B_1 = 90$ Show the case of the other angle similiarly.

Sorry for reviving an old thread, I just felt the urge to post this solution, and by the looks of things, it hasn't been posted in this thread. So here goes: Let $\gamma_1$ denote the circumcircle of triangle $A_1BD$, centred at the midpoint $M$ of $A_1B$ with radius $MB=MA_1$, since $\angle A_1DB=90^0$. Now, note that $JC_1E \sim JDC_1$ by trivial angle chasing, hence $JC_1^2=JE \cdot JD$. We note that the power of $J$ with respect to $\gamma_1$ is: $JM^2-MB^2=JA_1^2+A_1M^2-MB^2=JA_1^2=JC_1^2=JD \cdot JE$ implying that $E$ lies on $\gamma_1$, or $A_1EBD$ is cyclic, which readily implies that $\angle BEA_1=180^0-\angle BDA_1=\boxed{90^0}$. Similarly, let $\gamma_2$ denote the circumcircle of triangle $ADB_1$, again, since $\angle ADB_1=90^0$, we have that $\gamma_2$ is centred at $N$, the midpoint of $AB_1$, with radius $NA=NB_1$. Again, we have that the power of $J$ with respect to $\gamma_2$ is: $JN^2-NA^2=JB_1^2+B_1N^2-NA^2=JB_1^2=JC_1^2=JD \cdot JE$ implying that $E$ also lies on $\gamma_2$ i.e. $ADEB_1$ is cyclic i.e. $\angle AEB_1=\angle ADB_1=\boxed{90^0}$. Done!

We use directed angles modulo $180^{\circ}$. $A_1B_1\perp AB$ implies that $CA = CB$; let $M$ be the midpoint of $AB$, then let $AB = 2a, CM = b, AC = c$, so that $a^2+b^2 = c^2$. Observe that $DB = \frac ac(c-a)$, since $BA_1 = c-a$, hence $MD = \frac {a^2}{c}$, and $DC_1 = \frac{(c+a)(c-a)}{c} = \frac{b^2}{c}$. Hence, $$CD^2 - DC_1^2 = b^2 + \frac{a^4}{c^2} - \frac{b^4}{c^2} = b^2 + (a^2-b^2) = a^2 = c^2-b^2 = CJ^2 - C_1J^2$$And so $DJ\perp CC_1$. Thus $E$ is also the foot from $C$ to $DJ$. By Miquel's Theorem, $(CAB_1), (BDA_1), (ADB_1)$ concur at some point $E'$. Observe that $(CAB_1)$ is the circle with diameter $CJ$. We see that since $$\measuredangle A_1E'J = \measuredangle A_1CJ = \measuredangle CBA = \measuredangle A_1BD = \measuredangle A_1E'D$$Hence $D,J, E'$ are collinear. But $E'$ is on the circle with diameter $CJ$, so $CE'\perp E'J = DJ$, so $E'$ is the foot from $C$ to $DJ$, so $E' = E$. Therefore, since $E'$ is on $(A_1DB) = (A_1B)$, and $(ADB_1) = (AB_1)$, we see that $\measuredangle BEA_1 = \measuredangle AEB_1 = 90^{\circ}$.

My solution: Let $\angle BCA=2x,$ then $\angle AB_1D=x=\angle CA_1B_1=90-\angle BAC=90-\angle A_1B_1J.$ $1:$ $AC=BC.$ Proof: We know $\angle BAC=90-x$ and $\angle ACB=2x\to \angle ABC=180-\angle ACB-\angle BAC=90-x,$ as desired. $2:$ $EJB_1A_1$ is cyclic. Proof: We know $DC_1$ tangent to $(C_1EJ),$ since $\angle JC_1D=C_1EJ=90.$ Then $DC_1^2=DE\cdot DJ,(i)$ Also we have from PoP $DC_1^2=DA_1\cdot DB_1 (ii).$ From $(i),(ii)$ we have $DE\cdot DJ=DA_1\cdot DB_1\to EJB_1A_1$ is cyclic. Claim1: $BDA_1E$ is cyclic. Proof: $$\angle DBA_1=\angle ABC=^{1} \angle BAC=90-\angle AB_1D=\angle JB_1A_1=^{2} \angle DEA_1\to BDA_1E$$is cyclic.Then $\angle BEA_1+\angle BDA_1=\angle BEA_1+90=180\to \angle BEA_1=90.$ Claim 2: $ADA_1B_1$ is cyclic. Proof: $$\angle BAC=90-\angle DB_1A=\angle A_1B_1J=\angle DEA_1=^{2} 90-x,$$also $\angle A_1EB_1=2x\to \angle DEB_1=90+x,$ Also $\angle DAB_1=90-x\to \angle DAB_1+\angle DEB_1=180.$ Then $\angle ADB_1=\angle AEB_1=90.$ Then $\angle AEB_1=\angle BEA_1=90.$

Pretty easy. pohoatza wrote: In triangle $ABC$, let $J$ be the center of the excircle tangent to side $BC$ at $A_{1}$ and to the extensions of the sides $AC$ and $AB$ at $B_{1}$ and $C_{1}$ respectively. Suppose that the lines $A_{1}B_{1}$ and $AB$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $DJ$. Determine the angles $\angle{BEA_{1}}$ and $\angle{AEB_{1}}$. Proposed by Dimitris Kontogiannis, Greece We claim that $\boxed{\angle BEA_1 = \angle AEB_1 = 90^{\circ}}$. Denote by $\alpha = \angle EJC_1$. Now, we proceed with the following lemmas: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(18cm); 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 = -18.321507715121953, xmax = 21.63799503609756, ymin = -17.120184973658557, ymax = 10.568017192195112; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-4.034341463414636,7.532536585365856)--(-6.223042349268295,1.926256216585353), linewidth(0.8) + wrwrwr); draw((-6.223042349268295,1.926256216585353)--(2.4187186263414615,1.7879880409755966), linewidth(0.8) + wrwrwr); draw((2.4187186263414615,1.7879880409755966)--(-4.034341463414636,7.532536585365856), linewidth(0.8) + wrwrwr); draw((-4.034341463414636,7.532536585365853)--(xmin, 2.561464842005529*xmin + 17.866360404547578), linewidth(0.8) + wrwrwr); /* ray */ draw((-4.034341463414636,7.532536585365855)--(xmax, -0.8902053389382559*xmax + 3.94114427553417), linewidth(0.8) + wrwrwr); /* ray */ draw(circle((-0.7214002819942982,-6.267599593842906), 8.104792190337474), linewidth(1.2) + dotted + wrwrwr); draw((xmin, -0.3898063121755744*xmin + 1.605491316180467)--(xmax, -0.3898063121755744*xmax + 1.605491316180467), linewidth(0.8) + wrwrwr); /* line */ draw((-5.5097848482441645,3.753240228755382)--(-0.7214002819942982,-6.267599593842906), linewidth(0.8) + wrwrwr); draw((-8.271239983865076,-3.320130013913188)--(-3.2714474963483178,-0.9310159670358613), linewidth(0.8) + wrwrwr); draw((-6.223042349268295,1.926256216585353)--(-3.2714474963483178,-0.9310159670358613), linewidth(0.8) + wrwrwr); draw((-3.2714474963483178,-0.9310159670358613)--(-0.5917402023770499,1.836155382235093), linewidth(0.8) + wrwrwr); draw((-4.034341463414636,7.532536585365856)--(-3.2714474963483178,-0.9310159670358613), linewidth(0.8) + wrwrwr); draw((-3.2714474963483178,-0.9310159670358613)--(4.6675809392838925,-0.21396119654279033), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((-4.034341463414636,7.532536585365856),dotstyle); label("$A$", (-3.9070504078048804,7.871787767804867), NE * labelscalefactor); dot((-6.223042349268295,1.926256216585353),dotstyle); label("$B$", (-6.084774173658539,2.271926655609743), NE * labelscalefactor); dot((2.4187186263414615,1.7879880409755966),dotstyle); label("$C$", (2.556986801951217,2.133658479999987), NE * labelscalefactor); dot((-0.7214002819942982,-6.267599593842906),linewidth(4pt) + dotstyle); label("$J$", (-0.5886141931707338,-5.989596837073187), NE * labelscalefactor); dot((-0.5917402023770499,1.836155382235093),linewidth(4pt) + dotstyle); label("$A_{1}$", (-0.45034601756097775,2.099091436097548), NE * labelscalefactor); dot((4.6675809392838925,-0.21396119654279033),linewidth(4pt) + dotstyle); label("$B_{1}$", (4.803844655609754,0.059635845853644766), NE * labelscalefactor); dot((-8.271239983865076,-3.320130013913188),linewidth(4pt) + dotstyle); label("$C_{1}$", (-8.124229763902441,-3.0513981053658688), NE * labelscalefactor); dot((-5.5097848482441645,3.753240228755382),linewidth(4pt) + dotstyle); label("$D$", (-5.358866251707319,4.034845894634135), NE * labelscalefactor); dot((-3.2714474963483178,-0.9310159670358613),linewidth(4pt) + dotstyle); label("$E$", (-3.1465754419512217,-0.6662720760975751), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $ $ Lemma 1: $\overline{JA_{1}}$ is tangent to $\odot (ADB)$ and $\overline{JB_1} $ is tangent to $\odot (ADB_1)$. Proof. Since $A_1$ is the ex-touch point of $\overline{BC}$, it follows that $JA_1 \perp BC \implies \angle JA_1B = 90^{\circ} = \angle A_1DB$. Hence the result follows. Let $\gamma = \angle ACB$. Then by an easy angle chase we have that $$\angle JB_1D = \angle B_1AD = 90^{\circ}-\frac{\gamma}{2}$$Thus the result follows. $ $ Lemma 2: Quadrilaterals $A_1DBE$ and $ADEB_1$ are cyclic. Proof . Note that in $\triangle C_1EJ$, by sine rule we have $$JE = JC_1 \cdot \cos{\alpha}$$Similarly in $\triangle JC_1D$ we have $$ JD = \frac{JC_1}{\cos{\alpha}}$$So, $$JE \cdot JD = JC_1^2 = JA_1^2 = \text{Pow} \{J , \odot(A_1DB) \} \iff E \in \odot(A_1DB)$$Analogously we get $E \in \odot(ADB_1)$. So from both these cyclic quadrilaterals we get that: $$ \angle BEA_1 = 180^{\circ} - \angle A_1DB = 90^{\circ} = \angle ADB_1 = \angle AEB_1 \blacksquare$$

[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A1 = dir(15), B1 = dir(345), C1 = dir(90), J = origin, C = extension(A1,A1+dir(foot(B1,A1,J)--B1),J,J+dir(0)), B = extension(A1,C,C1,C1+dir(0)), A = extension(B,C1,C,B1), D = extension(B,C1,A1,B1), E = foot(C1,D,J); dot("$A_1$", A1, dir(15)); dot("$B_1$", B1, dir(345)); dot("$C_1$", C1, dir(90)); dot("$J$", J, dir(225)); dot("$C$", C, dir(330)); dot("$B$", B, dir(90)); dot("$A$", A, dir(45)); dot("$D$", D, dir(90)); dot("$E$", E, dir(180)); draw(arc(J,1,315,480)); draw(A--J--B1--A--C1--J--D--B1^^C1--E^^B--C--J); draw(B--E--A1^^A--E--B1, dashed); [/asy][/asy] We claim that $\boxed{\angle BEA_1 = \angle AEB_1 = 90^\circ}$. First, note that $$\angle A + \dfrac{1}{2}\angle C = \angle B_1AD + \angle AB_1D = \angle A_1BD + \angle BA_1D = \angle B + \dfrac{1}{2}\angle C,$$so $AC = BC$, or $\overline{AB} \parallel \overline{CJ}$. Now we use complex numbers. Set the $A$-excircle as the unit circle (so $J$ is the origin) with $c_1 = i$. Since $C$ lies on the real axis, $b_1 = \overline{a_1}$. It follows from the tangent formula that $$a = \dfrac{2b_1c_1}{b_1+c_1} = \dfrac{2i \cdot \tfrac{1}{a_1}}{\tfrac{1}{a_1}+i} = \dfrac{2i}{ia_1+1} = \dfrac{2}{a_1-i}$$and $$b = \dfrac{2a_1c_1}{a_1+c_1} = \dfrac{2ia_1}{a_1+i}.$$To solve for $d$, note that $$d+\overline{d} = 2\text{Re}(d) = 2\text{Re}(a_1) = a_1 + \dfrac{1}{a_1}$$and $$d-\overline{d} = 2\text{Im}(d) = 2i.$$Adding gives $$2d = a_1 + \dfrac{1}{a_1} + 2i \implies d = \dfrac{a_1^2+2ia_1+1}{2a_1}.$$We can also calculate $$\overline{d} = \dfrac{(\tfrac{1}{a_1})^2-\tfrac{2i}{a_1}+1}{\tfrac{2}{a_1}} = \dfrac{a_1^2-2ia_1+1}{2a_1}.$$It remains to find $e$. Since $E \in \overline{DJ}$, we can let $e = kd$ for some $k \in \mathbb{R}$. $\overline{C_1E} \perp \overline{DJ}$ implies \begin{align*} 0 &= \dfrac{c_1-e}{d-j} + \overline{\left(\dfrac{c_1-e}{d-j}\right)} \\ &= \dfrac{i-kd}{d} + \dfrac{-i-k\overline{d}}{\overline{d}} = \dfrac{i}{d} - k-\dfrac{i}{\overline{d}}- k \\ \implies 2k &= \dfrac{i}{d} - \dfrac{i}{\overline{d}} \\ \implies e = kd &= \dfrac{i}{2}\left(1-\dfrac{d}{\overline{d}}\right) \\ &= \dfrac{i}{2}\left(1-\dfrac{a_1^2+2ia_1+1}{a_1^2-2ia_1+1}\right) \\ &= \dfrac{2a_1}{a_1^2-2ia_1+1}.\end{align*}Finally, we verify \begin{align*} \dfrac{b-e}{a_1-e} &= \dfrac{\tfrac{2ia_1}{a_1+i}-\tfrac{2a_1}{a_1^2-2ia_1+1}}{a_1-\tfrac{2a_1}{a_1^2-2ia_1+1}} \\ &= \dfrac{2i(a_1^2-2ia_1+1)-2(a_1+i)}{(a_1+i)(a_1^2-2ia_1+1)-2(a_1+i)} \\ &= \dfrac{2a_1(ia_1+1)}{(a_1^2+1)(a_1-i)} \\ &= \dfrac{2ia_1}{a_1^2+1} = -\dfrac{\tfrac{-2i}{a_1}}{\tfrac{1}{a_1^2}+1} \\ &= -\overline{\left(\dfrac{b-e}{a_1-e}\right)}\end{align*}and \begin{align*} \dfrac{a-e}{b_1-e} &= \dfrac{\tfrac{2}{a_1-i} - \tfrac{2a_1}{a_1^2-2ia_1+1}}{\tfrac{1}{a_1} - \tfrac{2a_1}{a_1^2-2ia_1+1}} \\ &= \dfrac{2a_1(a_1^2-2ia_1+1)-2a_1^2(a_1-i)}{(a_1-i)(a_1^2-2ia_1+1)-2a_1^2(a_1-i)} \\ &= \dfrac{2a_1(-ia_1+1)}{-(a_1-i)(a_1^2+2ia_1-1)} = \dfrac{-2ia_1(a_1+i)}{-(a_1-i)(a_1+i)^2} \\ &= \dfrac{2ia_1}{a_1^2+1} = -\dfrac{\tfrac{-2i}{a_1}}{\tfrac{1}{a_1^2}+1} \\ &= -\overline{\left(\dfrac{a-e}{b_1-e}\right)}.\end{align*}The proof is now complete. $\blacksquare$

pohoatza wrote: In triangle $ABC$, let $J$ be the center of the excircle tangent to side $BC$ at $A_{1}$ and to the extensions of the sides $AC$ and $AB$ at $B_{1}$ and $C_{1}$ respectively. Suppose that the lines $A_{1}B_{1}$ and $AB$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $DJ$. Determine the angles $\angle{BEA_{1}}$ and $\angle{AEB_{1}}$. Proposed by Dimitris Kontogiannis, Greece [asy][asy] import olympiad; import geometry; size(12cm); pair O=origin; pair I,A,B,C,B1,A1,E,J,C1,D,M; A=dir(210); B=dir(330); C=dir(90); M=dir(30); I=incenter(A,B,C); J=2*M-I; B1=foot(J,A,C); C1=foot(J,A,B); A1=foot(J,B,C); D=foot(B1,A,B); E=foot(C1,D,J); draw(segment(A,B1),magenta); draw(segment(A,C1),magenta); draw(segment(B,C),magenta); draw(segment(J,D),magenta); draw(segment(J,B1),magenta); draw(segment(J,C1),magenta); draw(segment(J,A1),magenta); draw(segment(B1,D),magenta); filldraw(circumcircle(A,B,C),invisible,purple); filldraw(circumcircle(E,A1,C),invisible,dashed); filldraw(circumcircle(E,D,B),invisible,dotted); filldraw(circumcircle(A,D,E),invisible,dotted); filldraw(circumcircle(A1,B1,C1),invisible,purple); dot(A^^B^^C^^J^^A1^^B1^^C1^^E^^D); label("$A$",A,NW); label("$B$",B,NW); label("$C$",C,NW); label("$D$",D,NW); label("$E$",E,NW); label("$A_1$",A1,SW); label("$B_1$",B1,N); label("$C_1$",C1,SW); label("$J$",J,NE); [/asy][/asy] Claim : $ABC$ is an isosceles triangle Proof : $A_1B_1\perp AB\implies 90^{\circ}=\angle{ABC}+\angle{BA_1D}=\angle{ABC}+\angle{B_1A_1C}=\angle{ABC}+\tfrac{\angle{ACB}}{2}\implies \angle{BAC}=\angle{ABC}\square$ Now we have $JB_1^2=JC_1^2=JE\cdot JD\implies JB_1$ is tangent to $\odot{B_1ED}$.Similarly $JA_1$ is also tangent to $\odot{A_1ED}$.Thus \[\angle{JA_1E}=\angle{JDA_1}=\angle{JDB}=\angle{JB_1E} \implies \odot{JB_1A_1E}\].Thus $\angle{A_1BD}=\angle{ABC}=\angle{JB_1A_1}=\angle{A_1ED}\implies E\in \odot{A_1DB}$ Similarly $E\in \odot{ADB_1}$ Thus $E : A_1B\to B_1A$.Thus $\angle{BEA_1}=\angle{AEB_1}=90^{\circ} \; \blacksquare$

Solution with Jeffrey Kwan: We first show $\angle BEA_1 = 90^{\circ}$. Note that $E$ is the inverse of $D$ with respect to $(J)$. Let $B^\ast$ and $A^\ast$ denote the inverses of $B$ and $A$. Since $\measuredangle BDA_1 = \measuredangle BB^\ast A_1 = 90^{\circ}$, those four points are cyclic. Since $JB^\ast \cdot JB = JE \cdot JD$, point $E$ lies on that circle too. Hence $\measuredangle BEA_1 = 90^{\circ}$. The proof that $\angle B_1EA = 90^{\circ}$ is similar; note that $\measuredangle ADB_1 = \measuredangle AEB_1 = 90^{\circ}$ and $JA \cdot JA^\ast = JE \cdot JD$. [asy][asy] size(7cm); pair J = origin; pair C_1 = dir(180); pair A_1 = dir(110); pair B_1 = dir(70); pair D = extension(A_1, B_1, C_1, C_1+N); pair E = foot(C_1, D, J); pair B = 2*A_1*C_1/(A_1+C_1); pair A = 2*B_1*C_1/(B_1+C_1); pair C = 2*A_1*B_1/(A_1+B_1); pair T = -C_1; pair F = D+T-C_1; draw(A--B_1, lightred); draw(B--C, lightred); draw(D--A, lightred); draw(D--J, blue); filldraw(unitcircle, invisible, blue); filldraw(C_1--D--F--T--cycle, invisible, blue); draw(C_1--E, lightblue); draw(B--E--A_1, deepgreen+dashed); pair Bs = midpoint(A_1--C_1); draw(A_1--C_1, grey+dotted); draw(B--J, grey+dotted); dot("$J$", J, dir(270)); dot("$C_1$", C_1, dir(C_1)); dot("$A_1$", A_1, dir(A_1)); dot("$B_1$", B_1, dir(B_1)); dot("$D$", D, dir(D)); dot("$E$", E, dir(270)); dot("$B$", B, dir(B)); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$B^\ast$", Bs, dir(190)); /* TSQ Source: !size(10cm); J = origin R270 C_1 = dir 180 A_1 = dir 110 B_1 = dir 70 D = extension A_1 B_1 C_1 C_1+N E = foot C_1 D J R270 B = 2*A_1*C_1/(A_1+C_1) A = 2*B_1*C_1/(B_1+C_1) C = 2*A_1*B_1/(A_1+B_1) T = -C_1 F = D+T-C_1 A--B_1 lightred B--C lightred D--A lightred D--J blue unitcircle 0.1 yellow / blue C_1--D--F--T--cycle 0.1 yellow / blue C_1--E lightblue // First batch B--E--A_1 deepgreen dashed B* = midpoint A_1--C_1 R190 A_1--C_1 grey dotted B--J grey dotted // Second batch A--E--B_1 deepgreen dashed A* = midpoint B_1--C_1 R0 B_1--C_1 grey dotted A--J grey dotted */ [/asy][/asy]

Nice and easy 2006 G5

First of all, we know that the polar of $C$ with respect to the excircle is $DA_1B_1$, hence $C$ lies on the polar of $D$. It is easy to see that $C_1$ and $E$ also lie on the polar of $D$. Hence, we find that $\measuredangle CEJ = 90^\circ$, which implies $A_1CB_1EJ$ is cyclic with diameter $CJ$. Now, note that $\overline{DA_1B_1} \parallel \overline{C_1J}$, so $\overline{C_1J}$ is tangent to $(A_1CB_1EJ)$. Additionally, if $I$ is the incenter, then we know $\measuredangle ICJ = 90^\circ$, and so combined with our previous result we find that $AC = BC$. Now, evaluating the power at $C_1$ gives $$C_1A \cdot C_1B = s(s - c) = \frac{s(s - b)(s - c)}{(s - a)} = r^2 \cdot \frac{s^2}{(s - a)^2} = R_A^2 = C_1J^2 = C_1E \cdot C_1C,$$hence $E$ lies on the circumcircle. Next, we claim that $DBA_1E$ is cyclic, which quickly implies $\measuredangle BEA_1 = 90^\circ$. We proceed with a directed angle chase: $$ \measuredangle DBE = \measuredangle ABE = \measuredangle ACE = \measuredangle B_1CE = \measuredangle B_1A_1E = \measuredangle DA_1E.$$In the same vein, we can show that $ADEB_1$ is cyclic which implies $\measuredangle AEB_1 = 90^\circ$.

Note that $JA_1^2 = JB_1^2 = JC_1^2 = JD \cdot JE$, so $\overline{JA_1}$ and $\overline{JB_1}$ are tangent to $(A_1DE)$ and $(B_1DE)$, respectively. Now $\angle JA_1E = \angle A_1DE = \angle B_1DE = \angle JB_1E$, so $E$ lies on $(JA_1CB_1)$. Remark that $ACB_1E$ is an isosceles trapezoid, so $\angle B_1DE = \angle JB_1E = \angle JA_1E = 90^\circ - \angle A_1EB_1 = 90^\circ - \angle CB_1E = 90^\circ - \angle AB_1E = \angle B_1AE$, and thus $ADEB_1$ is cyclic. Now $\angle BAE = \angle DAE = \angle DB_1E = \angle A_1B_1E = \angle A_1CE = \angle BCE$, so $ABEC$ is cyclic. Finally, $\angle EDA_1 = \angle EDB_1 = \angle EAB_1 = \angle EAC = \angle EBC = \angle EBA_1$, so $EBDA_1$ is cyclic as well. Now an angle extraction on cyclic quadrilaterals $BDA_1E$ and $ADEB_1$ yields $\angle AEB_1 = 90^\circ$ and $\angle BEA_1 = 90^\circ$.

Can someone write a barycentric Solution, please because it looks good in bary but I can't finish my answer Thanks

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(120); pair B = dir(200); pair C = dir(340); pair J = 2 * dir(270) - incenter(A, B, C); pair A1 = foot(J, B, C); pair B1 = foot(J, A, C); pair C1 = foot(J, B, A); pair D = extension(A, B, A1, B1); pair E = foot(C1, D, J); draw(circumcircle(A, B, C)); draw(B1--A--C1); draw(B--C); draw(CP(J, A1)); draw(B1--D); draw(D--J); draw(C1 -- E); draw(C--J); draw(circumcircle(D,B,A1), dashed); draw(circumcircle(D, E, B1), dashed); dot('$A$', A, dir(A)); dot('$B$', B, dir(B)); dot('$C$', C, dir(C)); dot('$J$', J, dir(J)); dot('$A_1$', A1, dir(A1)); dot('$B_1$', B1, dir(B1)); dot('$C_1$', C1, dir(C1)); dot('$E$', E, dir(E)); dot('$D$', D, dir(D)); [/asy][/asy] Claim: $A_1, E, B, D$ is cyclic. Proof. Since $\angle BDA_1 = 90^{\circ},$ we have that the circumcenter $O$ of $\triangle BDA_1$ is on $B A_1,$ so then $\angle OA_1J = \angle BA_1J = 90^{\circ},$ implying that the two circles are orthogonal. Finally, $E, D$ are swapped by the inversion because $D$ is the pole of $C_1E,$ so $A_1, E, B, D$ is cyclic as claimed. $\blacksquare$ By symmetry, we also have $ABEB_1$ cyclic. Now, we know that $BDA_1, ADB_1 = 90^{\circ},$ so then $\angle BEA_1 = \angle AEB_1 = 90^{\circ}.$

Because of the perpendicularity, $(A_1DB)$ and $(B_1DA)$ are tangent to lines $A_1J$ and $B_1J$ respectively. Since $\triangle{JC_1D}$ is right, $JE \cdot JD = JC_1^2 = JB_1^2 = JA_1^2$, so by Power of a Point $A_1DBE$ and $B_1ADE$ are cyclic, which implies that $\angle{BEA_1} = \angle{A_EB_1} = 90^{\circ}$.

[asy][asy] size(250); pair A = (2, 6), B = origin, C = (9.84, 0); pair I = incenter(A, B, C); pair X1 = extension(circumcenter(A, B, C), (B+C)/2, (50, -50), (100, -50)); pair X2 [] = intersectionpoints(circumcenter(A, B, C)--X1, circumcircle(A, B, C)); pair L = X2[0]; pair J = 2*L - I; pair A1 = foot(J, B, C), B1 = foot(J, A, C), C1 = foot(J, A, B); pair D = foot(A1, A, B); pair EE = foot(C1, D, J); draw(A--B--C--cycle, red); draw(C1--A--B1, red); draw(C--C1, blue); draw(circumcircle(A1, B1, C1), magenta); draw(D--J, blue); draw(B1--D, orange); draw(circumcircle(B, EE, A1), green); draw(circumcircle(A1, B1, EE), green); draw(circumcircle(A, D, EE), green); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, NE); dot("$J$", J, S); dot("$A_1$", A1, S); dot("$B_1$", B1, NE); dot("$C_1$", C1, NW); dot("$E$", EE, S); dot("$D$", D, NW); [/asy][/asy] The answer is $90^{\circ}$ for both parts. Indeed, the entire problem is just equivalent to the three following claims: Claim. $\triangle ABC$ is isosceles. Proof. Simply notice that $$90^{\circ} - A = \angle CB_1D = \frac C2. \ \blacksquare$$ Claim. $DBEA_1$ is cyclic. Proof. We will show that $\angle DEA_1 = \angle B$. First, verify that quadrilateral $CA_1EJ$ is cyclic because $\angle CEJ = \angle CA_1J$, while $\triangle CA_1B_1$ is isosceles, so $$\angle DEA_1 = 90^{\circ} - \angle A_1EC = 90^{\circ} - \angle A_1B_1C = 90^{\circ} - \frac C2 = B,$$proving the claim. $\blacksquare$ Claim. $ADEB_1$ is cyclic. Proof. We will show that $\angle DEB_1 = 180^{\circ} - \angle A$. Observe that this is equivalent to $$\angle CEB_1 = 90^{\circ} - A \iff \angle CA_1B_1 = 90^{\circ} - A,$$which is obvious. $\blacksquare$ It is not hard to see that from these two claims, we must have $\angle BEA_1 = \angle AEB_1 = 90^{\circ}$, as desired. $\square$ Remark. I believe there are no nontrivial configuration issues in this problem -- if there are, they should not be hard to clean up.

miQuel Note that $\angle C_1JC=\angle C_1JA_1+\angle A_1JC=\angle DBA_1+\angle CA_1B_1=90^\circ.$ Let $F$ be $CJ\cup A_1B_1$ then $C_1DFJ$ rectangle. $\angle EJF=\angle DJF=\angle C_1DJ=\angle EC_1J$ so $C_1,E,C$ collinear. $\angle JEC=90^\circ.$ Thus, $EA_1CB_1J$ is cyclic. $\angle BAC=\angle JCB_1=\angle JEB_1$ so $ADEB_1$ is cyclic. Thus, $E$ is the miquel point of $ADA_1C$ which implies $\angle BEA_1=\angle AEB_1=90^\circ.$

Just a remark, is it just a coincidence or did ABC always seem isosceles? IDK but that looked like the only possible diagram Also a very trivial problem, not sure why G5 Anyways, here are two incredibly quick solutions I found in <15 minutes (even though they're short, it's easily motivated and also I think the first one is unique): 1. DEB_1=90+CEB_1=90+CJB_1=180-JCB_1. This is equal to 180-A iff CJ parallel to AB, which is evident since CJ perp. B_1A_1 perp. AB. Hence ADEB_1 is cyclic and AEB_1=90 degrees. Then from BA_1E=180-EA_1C=DJC=BDE, BDA_1E is cyclic so BEA_1=90 degrees as well. $\blacksquare$ 2. We have that =JB_1^2=JA_1^2=JC_1^2=JE*JD (similarity of right triangles and cross multiplying). Noting that JA_1 and JB_1 are tangents to A_1DB and ADB_1, this implies E lies on (A_1DB) and (ADB_1), so the supplement of right angles is the answer, as desired. $\blacksquare$

Note that the condition essentially says that the external bisector of $\angle C$ is parallel to $AB$, which is equivalent to $AC=BC$, so $\triangle ABC$ is isosceles. Claim; $C$, $E$, and $C_1$ are collinear. Note that $C_1E$ is the polar of $D$ with respect to the excircle, since $C_1$ clearly lies on the excircle and $E$ also does since it is the foot from $C_1$ to $DJ$. However, $D$ lies on the polar of $C$ (which is $A_1B_1$), so $C$ lies on the polar of $D$, showing the claim. Hence, since $$\angle CEJ=\angle CA_1J=90,$$$CA_1EJ$ is cyclic. Then, since $$\angle BA_1E=\angle EJC=\angle BDE,$$$BDA_1E$ is cyclic as well, so $\angle BEA_1=90$. Note that $B_1$ also lies on $(A_1CJE)$ since $\angle JB_1C=90$. Thus, $$\angle AB_1E=\angle BA_1E=\angle BDE,$$so $ADEB_1$ is cyclic and thus $\angle AEB_1=90$ as well. remark: lesson here, actually draw the circles :skull: solved it pretty quickly after i drew the excircle ok i really need to :sleeeeeeeeppppppp: now