We have : $ \angle{CAE}+\angle {CAI}+\angle {AIE}=\pi$ From $ AE||BE \rightarrow \angle{CAE}=\angle{ACB}$ Therefore $ \angle{AIE}=\frac{\angle{B}-\angle{C}}{2}$ Other $ \angle{APT}=\frac{\angle{B}-\angle{C}}{2}$ It gives $ \angle{AIE}=\angle{APE}$ so $ A,P,I,E$ lie on a circle . Because $ \angle{IEA}=\frac{\pi}{2}$ so $ \angle{API}=\frac{\pi}{2}$ From condition $ \angle{APB}+\angle{C}=\pi$ and $ PBI=\angle{B}$ we have $ \angle{A}=\frac{\pi}{2}$

a small error....the first eqn. RHS is equal to pi/2...

Iran TST 2008 P10 wrote: In the triangle $ ABC$, $ \angle B$ is greater than $ \angle C$. $ T$ is the midpoint of the arc $ BAC$ from the circumcircle of $ ABC$ and $ I$ is the incenter of $ ABC$. $ E$ is a point such that $ \angle AEI=90^\circ$ and $ AE\parallel BC$. $ TE$ intersects the circumcircle of $ ABC$ for the second time in $ P$. If $ \angle B=\angle IPB$, find the angle $ \angle A$. Solution: Let $\odot (AIE)$ $\cap$ $\odot (ABC)$ $=$ $D$. Let $AA'||BC$ with $A' \in \odot (ABC)$. [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 = -11.831784745493193, xmax = 17.492591423028003, ymin = -10.676873790658856, ymax = 6.749497293093495; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-4.06,4.19)--(-6.6,-5.07)--(4.68,-4.99)--cycle, linewidth(1.2) + rvwvcq); /* draw figures */ draw((-4.06,4.19)--(-6.6,-5.07), linewidth(1.2) + rvwvcq); draw((-6.6,-5.07)--(4.68,-4.99), linewidth(1.2) + rvwvcq); draw((4.68,-4.99)--(-4.06,4.19), linewidth(1.2) + rvwvcq); draw(circle((-0.9840987783166554,-1.6320722573515867), 6.5846559283971775), linewidth(0.8) + linetype("4 4")); draw((-4.06,4.19)--(-0.9374002653172535,-8.216562590267243), linewidth(0.4)); draw((-6.6,-5.07)--(3.784841528296612,2.9082913896854476), linewidth(0.4)); draw((-7.334197787673028,0.10974755730987962)--(4.68,-4.99), linewidth(0.4)); draw(shift((-3.2892794840301782,1.1278104357156291))*xscale(3.1576914100888476)*yscale(3.1576914100888476)*arc((0,0),1,-75.872693601157,104.127306398843), linewidth(0.4)); draw(shift((-3.2892794840301782,1.1278104357156291))*xscale(3.1576914100888476)*yscale(3.1576914100888476)*arc((0,0),1,104.12730639884302,284.127306398843), linewidth(0.4)); draw((-4.06,4.19)--(2.00891459611709,4.233041947490193), linewidth(0.4)); draw((-6.165234649953677,2.4316113886784727)--(2.091802443366688,-7.454144514703174), linewidth(0.4)); draw((2.00891459611709,4.233041947490193)--(2.091802443366688,-7.454144514703174), linewidth(0.4)); draw((-2.5620696258493183,4.200623619674828)--(-2.518558968060357,-1.9343791285687424), linewidth(0.4)); /* dots and labels */ dot((-4.06,4.19),dotstyle); label("$A$", (-4.452617955545645,4.562187267408717), NE * labelscalefactor); dot((-6.6,-5.07),dotstyle); label("$B$", (-7.312946450671894,-5.364835156852966), NE * labelscalefactor); dot((4.68,-4.99),dotstyle); label("$C$", (5.161931607903928,-5.196580539492599), NE * labelscalefactor); dot((-0.9374002653172535,-8.216562590267243),dotstyle); label("$M_A$", (-1.5682530865107733,-9.042400364872428), NE * labelscalefactor); dot((-1.0307972913160572,4.95241807556407),dotstyle); label("$M_{BC}$", (-1.1115619822469187,5.547678597662299), NE * labelscalefactor); dot((2.00891459611709,4.233041947490193),dotstyle); label("$A'$", (2.4217849823207995,4.682369136951837), NE * labelscalefactor); dot((3.784841528296612,2.9082913896854476),dotstyle); label("$M_B$", (4.224513025467594,3.264223076343025), NE * labelscalefactor); dot((-7.334197787673028,0.10974755730987962),dotstyle); label("$M_C$", (-8.418619650468594,0.35582183339952933), NE * labelscalefactor); dot((-2.518558968060357,-1.9343791285687424),dotstyle); label("$I$", (-2.4335625472212348,-1.687269948833505), NE * labelscalefactor); dot((-6.165234649953677,2.4316113886784727),dotstyle); label("$D$", (-6.808182598590791,2.7834955981705467), NE * labelscalefactor); dot((-2.5620696258493183,4.200623619674828),dotstyle); label("$E$", (-2.1931988081349956,3.360368571977521), NE * labelscalefactor); dot((2.091802443366688,-7.454144514703174),dotstyle); label("$F$", (2.6140759735897907,-7.840581669441232), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Applying Reim's Theorem we have, if $DI$ $\cap$ $\odot (ABC)$ $=$ $F$ $\implies$ $EI$ $||$ $A'F$ $\implies$ $F$ is $A-$antipode and Applying Reim's Theorem again $\implies$ $EI$ $||$ $M_{BC}M_A$ $\implies$ $D \equiv P$. Now, $\angle IPB$ $=$ $90^{\circ}$ $=$ $\angle B$ $\implies$ $\angle A=90^{\circ}$

We can assume that triangle $ABC$ is inscribed in the unit circle on Argand Plane. Let $a,b,c\in\mathbb C$ be numbers, such that $a^2,b^2,c^2$ are vertices of our triangle (so $|a|=|b|=|c|=1$)and it's incenter is $-(ab+bc+ca)$. Then $t=bc$. $$AE||BC\implies\frac{a^2-e}{b^2-c^2}=\overline{\left(\frac{a^2-e}{b^2-c^2}\right)}\implies \overline e=\frac{a^2-e}{b^2c^2}+\frac{1}{a^2}$$Thus $$IE\perp AE\implies\frac{a^2-e}{e+ab+bc+ca}=\overline{\left(\frac{a^2-e}{e+ab+bc+ca}\right)}=\frac{a^2(a^2-e)}{a^2(a^2-e)+b^2c^2+abc(a+b+c)}$$it means $$e=\frac{a^4+b^2c^2+abc(b+c)-a^3(b+c)}{2a^2}$$. $P$ lies on the circumcircle of triangle $ABC$ so $|p|=1$ $$\frac{p-t}{e-t}=\frac{2a^2(p-bc)}{(a^2-bc)(a^2-ab-bc-ca)}$$Points $P,T,E$ lie on one line: $$\frac{p-t}{e-t}=\overline{\left(\frac{p-t}{e-t}\right)}\iff$$$$\frac{2a^2(p-bc)}{(a^2-bc)(a^2-ab-bc-ca)}=\frac{2a^2bc(p-bc)}{p(a^2-bc)(bc-a^2-ab-ac)}\iff$$$$p=\frac{a^2-bc-ab-ac}{bc-a^2-ab-ac}\cdot bc$$$ \angle B=\angle IPB$ and $ \angle B$ is greater than $ \angle C$ so we have directed angles equality: $$\frac{\frac{p+ab+bc+ca}{p-b^2}}{|\frac{p+ab+bc+ca}{p-b^2}|}=\frac{\frac{a^2-b^2}{c^2-b^2}}{|\frac{a^2-b^2}{c^2-b^2}|}$$square both sides and use the identity $|x|^2=x\cdot \overline x$ to get $$\frac{\frac{p+ab+bc+ca}{p-b^2}}{\overline{\left(\frac{p+ab+bc+ca}{p-b^2}\right)}}=\frac{\frac{a^2-b^2}{c^2-b^2}}{\overline{\left(\frac{a^2-b^2}{c^2-b^2}\right)}}$$Auxiliary calculations: $$\frac{p+ab+bc+ca}{p-b^2}=\frac{a(abc-(a+b+c)(ab+bc+ca))}{b(b+c)(a+b)(a-c)}$$$$\overline{\left(\frac{p+ab+bc+ca}{p-b^2}\right)}=\frac{b(abc-(a+b+c)(ab+bc+ca))}{a(b+c)(a+b)(-a+c)}$$so LHS is equal to $\frac{-a^2}{b^2}$ Again AUX $$\overline{\left(\frac{a^2-b^2}{c^2-b^2}\right)}=\frac{a^2-b^2}{c^2-b^2}\cdot\frac{c^2}{a^2}$$so RHS is equal to $\frac{a^2}{c^2}$. We gained $$\frac{-a^2}{b^2}=\frac{a^2}{c^2}$$but $|a|=1\implies a\neq 0$ hence $$b^2+c^2=0$$which means that circumcenter is the midpoint of $BC$.

AlastorMoody wrote: Iran TST 2008 P10 wrote: In the triangle $ ABC$, $ \angle B$ is greater than $ \angle C$. $ T$ is the midpoint of the arc $ BAC$ from the circumcircle of $ ABC$ and $ I$ is the incenter of $ ABC$. $ E$ is a point such that $ \angle AEI=90^\circ$ and $ AE\parallel BC$. $ TE$ intersects the circumcircle of $ ABC$ for the second time in $ P$. If $ \angle B=\angle IPB$, find the angle $ \angle A$. Solution: Let $\odot (AIE)$ $\cap$ $\odot (ABC)$ $=$ $D$. Let $AA'||BC$ with $A' \in \odot (ABC)$. [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 = -11.831784745493193, xmax = 17.492591423028003, ymin = -10.676873790658856, ymax = 6.749497293093495; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-4.06,4.19)--(-6.6,-5.07)--(4.68,-4.99)--cycle, linewidth(1.2) + rvwvcq); /* draw figures */ draw((-4.06,4.19)--(-6.6,-5.07), linewidth(1.2) + rvwvcq); draw((-6.6,-5.07)--(4.68,-4.99), linewidth(1.2) + rvwvcq); draw((4.68,-4.99)--(-4.06,4.19), linewidth(1.2) + rvwvcq); draw(circle((-0.9840987783166554,-1.6320722573515867), 6.5846559283971775), linewidth(0.8) + linetype("4 4")); draw((-4.06,4.19)--(-0.9374002653172535,-8.216562590267243), linewidth(0.4)); draw((-6.6,-5.07)--(3.784841528296612,2.9082913896854476), linewidth(0.4)); draw((-7.334197787673028,0.10974755730987962)--(4.68,-4.99), linewidth(0.4)); draw(shift((-3.2892794840301782,1.1278104357156291))*xscale(3.1576914100888476)*yscale(3.1576914100888476)*arc((0,0),1,-75.872693601157,104.127306398843), linewidth(0.4)); draw(shift((-3.2892794840301782,1.1278104357156291))*xscale(3.1576914100888476)*yscale(3.1576914100888476)*arc((0,0),1,104.12730639884302,284.127306398843), linewidth(0.4)); draw((-4.06,4.19)--(2.00891459611709,4.233041947490193), linewidth(0.4)); draw((-6.165234649953677,2.4316113886784727)--(2.091802443366688,-7.454144514703174), linewidth(0.4)); draw((2.00891459611709,4.233041947490193)--(2.091802443366688,-7.454144514703174), linewidth(0.4)); draw((-2.5620696258493183,4.200623619674828)--(-2.518558968060357,-1.9343791285687424), linewidth(0.4)); /* dots and labels */ dot((-4.06,4.19),dotstyle); label("$A$", (-4.452617955545645,4.562187267408717), NE * labelscalefactor); dot((-6.6,-5.07),dotstyle); label("$B$", (-7.312946450671894,-5.364835156852966), NE * labelscalefactor); dot((4.68,-4.99),dotstyle); label("$C$", (5.161931607903928,-5.196580539492599), NE * labelscalefactor); dot((-0.9374002653172535,-8.216562590267243),dotstyle); label("$M_A$", (-1.5682530865107733,-9.042400364872428), NE * labelscalefactor); dot((-1.0307972913160572,4.95241807556407),dotstyle); label("$M_{BC}$", (-1.1115619822469187,5.547678597662299), NE * labelscalefactor); dot((2.00891459611709,4.233041947490193),dotstyle); label("$A'$", (2.4217849823207995,4.682369136951837), NE * labelscalefactor); dot((3.784841528296612,2.9082913896854476),dotstyle); label("$M_B$", (4.224513025467594,3.264223076343025), NE * labelscalefactor); dot((-7.334197787673028,0.10974755730987962),dotstyle); label("$M_C$", (-8.418619650468594,0.35582183339952933), NE * labelscalefactor); dot((-2.518558968060357,-1.9343791285687424),dotstyle); label("$I$", (-2.4335625472212348,-1.687269948833505), NE * labelscalefactor); dot((-6.165234649953677,2.4316113886784727),dotstyle); label("$D$", (-6.808182598590791,2.7834955981705467), NE * labelscalefactor); dot((-2.5620696258493183,4.200623619674828),dotstyle); label("$E$", (-2.1931988081349956,3.360368571977521), NE * labelscalefactor); dot((2.091802443366688,-7.454144514703174),dotstyle); label("$F$", (2.6140759735897907,-7.840581669441232), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Applying Reim's Theorem we have, if $DI$ $\cap$ $\odot (ABC)$ $=$ $F$ $\implies$ $EI$ $||$ $A'F$ $\implies$ $F$ is $A-$antipode and Applying Reim's Theorem again $\implies$ $EI$ $||$ $M_{BC}M_A$ $\implies$ $D \equiv P$. Now, $\angle IPB$ $=$ $90^{\circ}$ $=$ $\angle B$ $\implies$ $\angle A=90^{\circ}$ How do you import a graph from Geogebra?

JustKeepRunning wrote: How do you import a graph from Geogebra? It can be exported from GeoGebra Classic. If you find it easier to draw figures on GeoGebra Geometry, then you can first save it, and reopen in GeoGebra Classic.