Clearly $BCED$ is an isosceles trapezoid (cyclic trapezoid), Hence $BD=EC \implies \angle DAB=\angle EAC$, But $\angle ADE=\angle ACE$ $$\implies \triangle AEC' \sim \triangle ABD \text{ (by AA test) } \implies \dfrac{AD}{AC'}=\dfrac{AB}{AE} \implies AD \cdot AE=AB \cdot AC'$$We have to show that $AD \cdot AE =AB \cdot AC'= AI^2$ Hence it suffice to show that $\triangle AC'I \sim \triangle AIB$ But notice that since $DE || BC \implies \angle AC'B'=\angle ACB$ We observe that there is homothety centered at $A$ with a certain ratio taking $B'C'$ to $BC$ Which infact means that the incentre of $\triangle ABC$ is the excentre of $\triangle AB'C' \implies CI$ is the angle bisector of the external angle $\angle B'C'I=180-C$ $$\implies \angle AC'I=90+\dfrac{C}{2} \implies \triangle AC'I \sim \triangle AIB \text{ (by AA test) }$$The above result follows from the fact that $AI$ is the angle bisector of $\angle BAC$, and $\angle AIB=90+\dfrac{C}{2}$

Simply screams inversion Sharygin P15 wrote: A line $l$ parallel to the side $BC$ of triangle $ABC$ touches its incircle and meets its circumcircle at points $D$ and $E$. Let $I$ be the incenter of $ABC$. Prove that $AI^2 = AD \cdot AE$. Define $M_A:= AI\cap (ABC).$ We begin with the following claim. Claim: $AI$ bisects $\angle DAE$ Proof: We know that $AM_A$ bisects $\angle BAC.$ Hence $M_AB=M_AC.$ Since $$DE||BC \implies M_AD=M_AE \implies AM_a \text{ bisects }\angle DAE $$ Define $B':= AC\cap DE, C':= AB\cap DE.$ Define $G:= (I)\cap DE, J:= (I)\cap BC, H:= (I)\cap AB.$ We state the following lemma, which is true for any tangential quad satisfying the fact that $I\in GH.$ Lemma: $\Delta AIB \sim \Delta AB'I$ Proof: Since $$C'B'||BC \implies \angle BC'I+\angle C'BI=90^{\circ}\implies \angle C'IB=90^{\circ}$$$$\implies \angle C'BI=\angle C'IH=\angle C'IG.$$Note that $I$ is the $A-$excentre in $\Delta AC'B'.$ Letting $\angle AB'C'=C, \angle AC'B'=B,$ we get $$\angle C'IA=180-\angle AC'I-\angle C'AI= 180-(\angle AC'B'+90-\angle AC'B'/2 )- A/2$$$$=180-(90+B/2)-A/2=C/2= 90-(90-C/2)=\angle GIB'$$$$\implies \angle AIB'=\angle CIG'=\angle ABI $$And note that $AI$ bisects $\angle BAC.$ Hence $$\Delta AIB \sim \Delta AB'I$$ Hence $AI^2=AB'\cdot AB$ and similarly $AI^2=AC'\cdot AC.$ Now we are ready to invert! Invert wrt $(A,AI)$ followed by reflection wrt $AI.$ Then it's enough to show that $D\leftrightarrow E.$ Now, note that $B\leftrightarrow B', C\leftrightarrow C' \implies (ABC)\leftrightarrow B'C'\implies DE\leftrightarrow (ABC).$ Now let's try to find out where $D$ will go. Note that $D^{*}$ ( the inverted point $D$) would lie on $(ABC).$ Moreover, $AI$ would also bisect $\angle DAD^{*}.$ But $E\in (ABC)$ and $AI \text { bisects } \angle DAD^{*}. $ Hence $D^{*}=E.$ Hence $AI^2=AD\cdot AE.$

Denote by $\omega$ and $\Omega$ the incircle and circumcircle of $\triangle ABC$, respectively. Denote by $\Gamma$ the circumcircle of $\triangle DIE$. Let $N$ be the midpoint of big arc $\widehat{BAC}$ of $\Omega$. Since $DE \parallel BC$, $N$ is also the midpoint of $\widehat{DE}$. [asy][asy] size(10cm); pair A = dir(120), B = dir(210), C = dir(-30), I = incenter(A, B, C), N = dir(90), D = intersectionpoints(circle(N, abs(N-I)), unitcircle)[0], E = intersectionpoints(circle(N, abs(N-I)), unitcircle)[1], T = 2*foot(origin, N, I)-N, I1 = 2A-I, E1 = 2*foot(E, A, N)-E; draw(A--B--C--cycle, blue); draw(unitcircle, royalblue); draw(incircle(A, B, C), royalblue); draw(circle(N, abs(N-I)), deepcyan); draw(D--E--T--cycle, purple); draw(T--N, blue); draw(D--E1^^I--I1, lightblue); string[] names = {"$A$", "$B$", "$C$", "$D$", "$E$", "$I$", "$N$", "$T$", "$I_1$", "$E_1$"}; pair[] points = {A, B, C, D, E, I, N, T, I1, E1}; pair[] ll = {dir(150), B, C, dir(190), dir(-10), dir(220), N, T, I1, E1}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] Claim. The center of $\Gamma$ is $N$. Proof. By Poncelet's Porism, the tangents from $D$ and $E$ to $\omega$ (other than $DE$ itself) meet on $\Omega$ at a point $T$. Now, $I$ is the incenter of $\triangle TED$ and $N$ is the arc midpoint of its circumcircle, so the claim follows by the Incenter-Excenter lemma. Let $IA$ meet $\Gamma$ again at $I_1 \neq I$. Since $NA \perp AI$, we have $AI = AI_1$. Let $DA$ meet $\Gamma$ again at $E_1 \neq D$. Since $AN$ is the exterior angle bisector of $\angle DAN$ (due to $N$ being the arc midpoint), it is the interior angle bisector of $\angle E_1AE$. Therefore, by symmetry, $AE=AE_1$. By power of a point with respect to $\Gamma$, we have \[ AI \cdot AI_1 = AD \cdot AE_1, \]which, combined with our established length equalities, implies the result.

Well, similarly as above but still I will post. We know that the line formed by intersections of the tangents from the mixtilinear touchpoint $T$ to the incircle with the circumcircle is tangent to the incircle (Poncelet Porism). Now we know $\angle DIE= 90^o+\angle DTE/2=180^o-\angle DAE/2$, so $\angle AIE=\angle ADI$, and similarly $\angle AID= \angle AEI$, so $AID$ and $AEI$ are similar, done.

remark that $AD,AE $ are isogonal ,consider then the function $f$ which is composition of the inversion with the power $AD.AE$ and the symmetry in the bisector $AI$ it maps $DE\to (ABC), (I)\to (K)$, the $A-$mixtlinear ; so if $(I),(K)$ touch resp. $AB,AC$ at $F,G$ and $V,U$ then $f$ swaps $ F\leftrightarrow U,G\leftrightarrow V$ thus $AD.AE=AF.AU=AF.AV$ but since $FIV $ is right angled and $AI\perp IV$ then the circle $(FVI)$ is tangent to $AI$ leads $AD.AE=AF.AV=AI^2$ RH HAS

Definitely overcomplicated this one.

This problem is essentially just proving that $TB \cdot TC = TI^2$, where $T$ is the $A$-mixtilinear touchpoint. As to why this is true, reflect $I$ over $T$ to get $I'$, and let the midpoints of the minor and major arcs $BC$ be $M$ and $S$ respectively; then since $\angle{ITM} = \angle{STM} = 90^\circ$, $BICI'$ is cylic, and moreover since $SB = SC$ are both tangent to $(BIC)$, $II'$ is the symmedial chord of $BIC$. So $T$ is the $I$-dumpty point of $BIC$.

Let $DE$ intersect $AC$ at $B'$. By inversion around $A$ with radius $\sqrt{AD\cdot AE}$,$B$ goes to $B'\Rightarrow AB\cdot AB' = AD \cdot AE$ Notice that triangles $ABI$ and $AIB'$ are similar which gives $AI^2=AB\cdot AB'=AD\cdot AE$ Q.E.D.

Infinityfun wrote: Let $DE$ intersect $AC$ at $B'$. By inversion around $A$ with radius $\sqrt{AD\cdot AE}$,$B$ goes to $B'\Rightarrow AB\cdot AB' = AD \cdot AE$ Notice that triangles $ABI$ and $AIB'$ are similar which gives $AI^2=AB\cdot AB'=AD\cdot AE$ Q.E.D. Why does $B$ go to $B^{'}$?

khina wrote: This problem is essentially just proving that $TB \cdot TC = TI^2$. Why?

This is very nice and interesting problem. I see a generalization with two isogonal conjugate points Let $P$ and $Q$ be two isogonal conjugate points with respect to triangle $ABC$. Let $K$ and $L$ be the orthogonal projections of $P$ and $Q$ on the line $BC$. Let $KN$ and $LM$ be the diameters of pedal circle of $P$ and $Q$. Assume that line $MN$ meet circumcircle of triangle $ABC$ at $E$ and $F$. Prove that $$AP\cdot AQ=AE\cdot AF.$$ Quote: My 2001th post.

howcanicreateacccount wrote: khina wrote: This problem is essentially just proving that $TB \cdot TC = TI^2$. Why? bump for questions

Post 4 is roughly what I'm going for.

Mathlover_1 wrote: Infinityfun wrote: Let $DE$ intersect $AC$ at $B'$. By inversion around $A$ with radius $\sqrt{AD\cdot AE}$,$B$ goes to $B'\Rightarrow AB\cdot AB' = AD \cdot AE$ Notice that triangles $ABI$ and $AIB'$ are similar which gives $AI^2=AB\cdot AB'=AD\cdot AE$ Q.E.D. Why does $B$ go to $B^{'}$? That is $\sqrt{BC}$ inversion so $DE$ invert to $(ABC)$.Let $B$ invert to $B^{'}$ , and $B^{'}$ must be on $AC$.We know that B lies on $(ABC)$ and $(ABC)$ invert to $DE$ so $B^{'}$ also lies on $DE$.Then $B^{'}$ must be intersection of $AC$ and $DE$

The inversion with centre $A$ and radius $AI$ maps the incircle to the $A$-mixtilinear incircle (because it maps the circle with diametre $AI$ to the line through $I$ that is perpendicular to $AI$ and hence it maps the incircle to a circle which is tangent to $AB$ and $AC$ at the points where the line through $I$ that is perpendicular to $AI$ intersects them, but said circle is the $A$-mixtilinear incircle), so it maps $(ABC)$ to a line tangent to the incircle. Hence the inversion with centre $A$ and radius $AI$ followed by the reflection wrt $AI$ maps $(ABC)$ to a line tangent to the incircle and parallel to $BC$, which is not $BC$ (because this isn't $\sqrt{bc}$-inversion) and is therefore $DE$. Hence this transformation maps $D$ to $E$, which means that $AI^2=AD\cdot AE$.