Spiral similarity

Basic information

A spiral similarity is a plane transformation composed of a rotation of the plane and a dilation of the plane having the common center. The order in which the composition is taken is not important.

Any two directly similar figures are related either by a translation or by a spiral similarity (directly similar figures are similar and have the same orientation).

The transformation is linear and transforms any given object into an object homothetic to given.

On the complex plane, any spiral similarity can be expressed in the form $T(x) = x_0+k (x-x_0),$ where $k$ is a complex number. The magnitude $|k|$ is the dilation factor of the spiral similarity, and the argument $\arg(k)$ is the angle of rotation.

The spiral similarity is uniquely defined by the images of two distinct points. It is easy to show using the complex plane.

Let $A' = T(A), B' = T(B),$ with corresponding complex numbers $a', a, b',$ and $b,$ so $a' = T(a) = x_0 + k (a - x_0), b' = T(b) = x_0+ k (b-x_0) \implies$ $k = \frac {T(b) - T(a)}{b-a} = \frac {b' - a' }{b - a},$ $x_0=\frac {ab' - ba' }{a-a'+b' -b}, a' - a \ne b' - b.$ For any points $A, B, A',$ and $B'$ the center of the spiral similarity taking $AB$ to $A'B'$ point $x_{0}$ is also the center of a spiral similarity taking $A'B$ to $AB'.$ This fact explain existance of Miquel point.

Case 1 Any line segment $AB$ can be mapped into any other $A'B'$ using the spiral similarity. Notation is shown on the diagram. $P = AB \cap A'B'.$

$\Omega$ is circle $AA'P, \omega$ is circle $BB'P, x_0 = \Omega \cap \omega, x_0 \neq P,$

$C$ is any point of $AB, \theta$ is circle $CPx_0, C' = \theta \cap A'B'$ is the image $C$ under spiral symilarity centered at $x_0.$ $\triangle AA'x_0 \sim \triangle BB'x_0 \sim \triangle CC'x_0.$

$|k| = \frac {A'B'}{AB} = \frac {A'x_0}{Ax_0} = \frac {B'x_0}{Bx_0} = \frac {C'x_0}{Cx_0}$ is the dilation factor,

$\arg(k) =\angle APA'=\angle Ax_0A' =\angle Bx_0B' =\angle Cx_0C'$ is the angle of rotation.

Case 2 Any line segment $AB$ can be mapped into any other $BB'$ using the spiral similarity. Notation is shown on the diagram. $B = AB \cap BB', \Omega$ is circle $ABB$ (so circle is tangent to $BB'), \omega$ is circle tangent to $AB, x_0 = \Omega \cap \omega, x_0 \neq B, C$ is any point of $AB, \theta$ is circle $CBx_0,$ $C' = \theta \cap BB'$ is the image $C$ under spiral symilarity centered at $x_0.$ $\triangle ABx_0 \sim \triangle BB'x_0 \sim \triangle CC'x_0.$ $|k| = \frac {BB'}{AB}$ is the dilation factor,

$\angle Ax_0B = \arg(k)$ is the angle of rotation.

Simple problems

Explicit spiral similarity

Given two similar right triangles $ABC$ and $A'B'C, k = \frac {AC}{BC},$ $\angle ACB = 90^\circ, D = AA' \cap BB'.$ Find $\angle ADB$ and $\frac {AA'}{BB'}.$

Solution

The spiral similarity centered at $C$ with coefficient $k$ and the angle of rotation $90^\circ$ maps point $B$ to point $A$ and point $B'$ to point $A'.$

Therefore this similarity maps $BB'$ to $AA' \implies$ $\frac {AA'}{BB'} = k, \angle ADB = 90^\circ.$

Hidden spiral symilarity

Let $\triangle ABC$ be an isosceles right triangle $(AC = BC).$ Let $S$ be a point on a circle with diameter $BC.$ The line $\ell$ is symmetrical to $SC$ with respect to $AB$ and intersects $BC$ at $D.$ Prove that $AS \perp DS.$

Proof

Denote $\angle SBC = \alpha, \angle SCB = \beta = 90^\circ - \alpha,$ $\angle SCA = \alpha, \angle BSC = 90^\circ, k = \frac {SC}{SB} = \cot \beta.$ Let $SC$ cross perpendicular to $BC$ in point $B$ at point $D'.$

Then $\frac {BC}{BD'} = \cot \beta.$

Points $D$ and $D'$ are simmetric with respect $AB,$ so $BD = BD' \implies k = \frac {SC}{SB} = \frac {BC}{BD}.$

The spiral symilarity centered at $S$ with coefficient $k$ and the angle of rotation $90^\circ$ maps $B$ to $C$ and $D$ to point $D_0$ such that $k \cdot BD_0 = BC = AC, \angle D_0CS = \angle DBS \implies D_0 = A.$

Therefore $\angle ASC = \angle DSB \implies$ $\angle ASD = \angle ASC - \angle DSC = \angle DSB - \angle DSC = \angle BSC = 90^\circ.$

Linearity of the spiral symilarity

$\triangle ABF \sim \triangle BCD \sim \triangle CAE.$ Points $D,E,F$ are outside $\triangle ABC.$

Prove that the centroids of triangles $\triangle ABC$ and $\triangle DEF$ are coinsite.

Proof

Let $\vec y = T(\vec x),$ where $T$ be the spiral similarity with the rotation angle $\angle BAF= \angle CBD = \angle ACE$ and $k = \frac {|AF|}{|AB|} = \frac {|DB|}{|BC|} = \frac {|EC|}{|CA|}.$

A vector has two parameters, modulo and direction. It is not tied to a center of the spiral similarity. Therefore $\vec AF = T(\vec AB), \vec BD = T(\vec BC), \vec CE = T(\vec CA).$ $\vec AB + \vec BC + \vec CA = \vec 0.$ We use the property of linearity and get $\vec AF + \vec BD + \vec CE = k(\vec AB + \vec BC + \vec CA) = k \cdot \vec 0 = \vec 0.$ Let $G$ be the centroid of $\triangle ABC$ so $\vec GA + \vec GB + \vec GC = \vec 0 \implies$

$\vec GD + \vec GE + \vec GF = \vec 0 \implies G$ is the centroid of the $\triangle DEF.$

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Construction of a similar triangle

Let triangle $\triangle ABC$ and point $A'$ on sideline $BC$ be given. Construct $\triangle A'B'C' \sim \triangle ABC$ where $B'$ lies on sideline $AC$ and $C'$ lies on sideline $AB.$

Solution

Let $T(X)$ be the spiral symilarity centered at $A'$ with the dilation factor $k = \frac {AB}{AC}$ and rotation angle $\alpha = \angle BAC, A_1 = T(A), B_1 = T(B).$

$A_1B_1 = T(AB)$ so image of any point $C' \in AB$ lies on $A_1B_1.$ $B' \in AC \implies B' = A_1B_1 \cap AC.$ The spiral symilarity $T^{-1}(X)$ centered at $A'$ with the dilation factor $k^{-1} = \frac {AC}{AB}$ and rotation angle $-\alpha$ maps $B'$ into $C'$ and $\angle B'A'C' = \alpha, \frac {A'C'}{A'B'} = k^{-1} = \frac {AC}{AB}$ therefore the found triangle $\triangle A'B'C' \sim \triangle ABC$ is the desired one.

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Center of the spiral symilarity for similar triangles

Let triangle $\triangle ABC$ and point $C' (C' \ne C, C' \ne B)$ on sideline $BC$ be given. $\triangle A'B'C' \sim \triangle ABC$ where $B'$ lies on sideline $AB$ and $A'$ lies on sideline $AC.$ The spiral symilarity $T$ maps $\triangle ABC$ into $\triangle A'B'C'.$ Prove

a) $\angle AB'A' = \angle BC'B' = \angle CA'C'.$

b) Center of $T$ is the First Brocard point of triangles $\triangle ABC$ and $\triangle A'B'C'.$

Proof

a) Let $\tau(X)$ be the spiral symilarity centered at $C'$ with the dilation factor $k = \frac {AC}{BC}$ and rotation angle $\gamma = \angle ACB, A_1 = \tau(A), B_1 = \tau(B).$ $A' = A_1B_1 \cap AC, B' = \tau^{-1}(A').$

Denote $\varphi = \angle BC'B' \implies \angle CC'A' = 180^\circ - \varphi - \gamma \implies$ $\angle CA'C' = 180^\circ - \gamma - \angle CC'A' = \varphi.$ Similarly $\angle AB'A' = \varphi.$

b) It is well known that the three circumcircles $AA'B', BB'C',$ and $CC'A'$ have the common point (it is $D$ in the diagram).

Therefore $AB'DA'$ is cyclic and $\angle AB'A' = \angle ADA' =\varphi.$

Similarly, $\angle BDB' = \angle CDC' = \varphi.$

$\angle A'DC' = 180^\circ - \gamma, \angle A'DC = 180^\circ - \gamma - \varphi \implies$ $\angle ADC = \angle A'DC' = 180^\circ - \gamma.$ $\angle CAD + \angle ACD = 180^\circ - \angle ADC = \gamma = \angle ACD + \angle BCD \implies \angle CAD = \angle BCD = \psi.$

Similarly, $\angle CAD = \angle ABD = \angle BCD = \psi.$

Therefore, $D$ is the First Brocard point of $\triangle ABC.$

$AB'DA'$ is cyclic $\implies \angle A'B'D = \angle A'AD = \psi.$ Similarly, $\angle B'C'D = \angle C'A'D = \psi.$

Therefore $D$ is the First Brocard point of $\triangle A'B'C',$ and $\triangle A'DC' \sim \triangle ADC, \triangle A'DB' \sim \triangle ADB.$

Therefore the spiral symilarity $T$ maps $\triangle ABC$ into $\triangle A'B'C'$ has the center $D,$ the angle of the rotation $\varphi.$

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Spiral similarity in rectangle

Let rectangle $ABCD$ be given. Let point $E \in AC, DE \perp AC.$

Let points $F$ and $G$ be the midpoints of segments $AE$ and $BC,$ respectively.

Prove that $DF \perp FG.$

Proof

Let $H$ be the midpoint $DE \implies FH = \frac {AD}{2} = CG,$

$FH || AD ||BC || CG \implies FHCG$ is a parallelogram $\implies CH || GF.$

$\angle DAE = \angle CDE, \angle AED = \angle DEC = 90^\circ \implies$ $\triangle AED \sim \triangle DEC \implies$ $DF$ and $CH$ are corresponding medians of $\triangle AED$ and $\triangle DEC.$

There is a spiral similarity $T$ centered at $E$ with rotation angle $90^\circ$ that maps $\triangle AED$ to $\triangle DEC.$ Therefore $T(DF) = CH \implies DF \perp CH \implies DF \perp FG.$

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Common point for 6 circles

Let $\triangle ABC$ and point $A'$ on sideline $BC$ be given. $\triangle A'B'C' \sim \triangle ABC$ where $B'$ lies on sideline $AC$ and $C'$ lies on sideline $AB.$

Denote $D = BB' \cap CC', E = AA' \cap CC', F = BB' \cap AA'.$

Prove that circumcircles of triangles $\triangle ABF, \triangle A'B'F, \triangle BCD,$ $\triangle B'C'D, \triangle ACE, \triangle A'C'E$ have the common point.

Proof

$\triangle A'B'C' \sim \triangle ABC$ so there is the spiral symilarity $T$ taking $\triangle ABC$ to $\triangle A'B'C', T(ABC) = A'B'C'.$ Denote $O$ the center of $T.$ $A'B' = T(AB), A'B = \tau(AB'), F = AA' \cap BB' \implies$ the center of $\tau$ is the secont crosspoint of circumcircles of $\triangle ABF$ and $\triangle A'B'F,$ but this center is point $O,$ so these circles contain point $O$ . Similarly for another circles.

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Three spiral similarities

Let triangle $\triangle ABC$ be given. The triangle $\triangle ACE$ is constructed using a spiral similarity of $\triangle ABC$ with center $A$ , angle of rotation $\angle BAC$ and coefficient $\frac {AC}{AB}.$

A point $D$ is centrally symmetrical to a point $B$ with respect to $C.$

Prove that the spiral similarity with center $E$ , angle of rotation $\angle ACB$ and coefficient $\frac {BC}{AC}$ taking $\triangle ACE$ to $\triangle CDE.$

Proof

$\angle ECD = 180^\circ - \angle ACB - \angle ABC = \angle BAC.$ $EC = BC \cdot \frac {AC}{AB}, CD = BC = AC \cdot \frac {BC}{AC} \implies \frac {CD}{EC}=\frac {AB}{AC} \implies \triangle CDE \sim \triangle ACE \implies \angle DEC = \angle CEA.$

Corollary

Three spiral similarities centered on the images of the vertices of the given triangle $\triangle ABC$ and with rotation angles equal to the angles of $\triangle ABC$ take $\triangle ABC$ to $\triangle FDC$ centrally symmetric to $\triangle ABC$ with respect to $C.$

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Superposition of two spiral similarities

Let $T$ be the spiral similarity centered at $A$ with angle $\alpha$ and coefficient $\frac {1}{|k|}, Y = T(X), \frac {XA}{YA} = k.$

Let $t$ be spiral similarity centered at $B$ with angle $\alpha$ and coefficient $k, Z = t(Y), \frac {BZ}{BY} = k.$

Let $t(T(O)) = O.$

Prove: a) $XO = ZO, AO = BO, O$ is the crosspoint of bisectors $AB$ and $XZ.$

b) $\angle XOZ = 2 \alpha, \angle AOB = 2 \angle AXY.$

Algebraic proof

We use the complex plane $x = \vec X, y = \vec Y, z = \vec Z, a = \vec A,b = \vec B, o = \vec O.$ $y = T(x) = a + \frac {e^{i \alpha}}{k}(x - a), z = t(y) = b + k e^{i \alpha}(y - b) \implies$ $z = e^{2 i \alpha}x + b(1 - k e^{i \alpha}) + a e^{i \alpha}(k - e^{i \alpha}) = o + e^{2 i \alpha}(x - o).$ Let $a = (-1,0), b = (1,0).$ Then $o = i(\cot \alpha - \frac {k}{\sin \alpha}) \implies XO = ZO, AO = BO, \angle XOZ = 2 \alpha, \angle AOB = 2 \angle AXY.$ Geometric proof

Denote $\angle XAY = \alpha, \angle AXY = \beta, \angle AYX = \gamma = 180^\circ - \alpha - \beta.$

Then $k = \frac {XA}{YA} = \frac {\sin \gamma}{\sin \beta}, \triangle BZY \sim \triangle AXY.$

Let $H$ be the midpoint $AB, O$ be the point on bisector $AB$ such that $\angle AOH = \beta, C$ be the point on bisector $AB$ such that $\angle OAC = \alpha.$ Then

$\triangle AOC = \triangle BOC \sim \triangle AXY \sim \triangle BZY \implies$ $k = \frac {OA}{CA}, k^{-1} = \frac {CB}{OB},\angle CBO =\alpha \implies$ $T(O) = C, t(C) = O, t(T(O)) = O, \angle AOB = 2 \beta.$ $T(XO) = YC \implies \frac {XO}{YC} = k; t(YC) = ZO \implies \frac {ZO}{YC} = k \implies$ $XO = ZO, \angle (XO)(YC) = \angle (YC)(ZO) = \alpha \implies \angle (XO)(ZO) = \angle XOZ = 2 \alpha.$ $XO = ZO, AO = BO \implies O$ is the crosspoint of bisectors $AB$ and $XZ.$

Corollary

There is another pare of the spiral similarities centered at $X$ and $Z$ with angle $\beta,$ coefficients $k' = \frac {XA}{XY} = \frac {\sin \gamma}{\sin \alpha}$ and $k'^{-1}, Y = T'(A), B = t'(Y).$

In this case $t'(T'(O)) = O.$

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Spiral similarity for circles

1. Let circle $\omega$ cross circle $\Omega$ at points $A$ and $B.$ Point $C$ lies on $\omega.$

Spiral similarity $T$ centered at $A$ maps $\omega$ into $\Omega: \Omega = T(\omega), C' = T(C).$

Prove that points $C, C',$ and $B$ are collinear.

Proof

Arcs $\overset{\Large\frown} {AC} = \overset{\Large\frown} {ABC'}, \angle ABC = \frac {\overset{\Large\frown} {AC}}{2},$ $\angle ABC' = \frac {360^\circ -\overset{\Large\frown} {ABC'}}{2} = 180 ^\circ - \frac {\overset{\Large\frown} {ABC'}}{2} = 180 ^\circ - \angle ABC' \implies$ $B \in CC'.$

Corollary

Let $C \in \omega, C' \in \Omega,$ points $C, C',$ and $B$ be collinear.

Then exist the spiral similarity $T$ centered at $A$ such that $T(\omega) = \Omega, T(C) = C'.$

2. Let circle $\omega$ cross circle $\Omega$ at points $A$ and $B.$

Points $C$ and $D$ lie on $\omega, C' = BC \cap \Omega, D' = BD \cap \Omega.$

Let $EA$ be the tangent to $\omega, FA$ be the tangent to $\Omega.$

Prove that angle between tangents is equal angle between lines $CD$ and $C'D'.$

Proof

There is the spiral similarity $T$ centered at $A$ such that $T(\omega) = \Omega, T(C) = C', T(D) = D'.$ Therefore $T(CD) = C'D', T(EA) = FA \implies$ angles between these lines are the same.

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Remarkable point for spiral similarity

Circles $\omega$ and $\Omega$ centered at points $O$ and $O',$ respectively intersect at points $A$ and $B.$ Points $C \in \omega, C' \in \Omega,$ and $A$ are collinear. Point $X$ is symmetrical to $A$ with respect to the midpoint $OO'$ point $M.$ Prove:

a) $\angle ABX = 90^\circ,$ b) $CX = C'X.$

Proof

a) $AB \perp OO', OO'$ cross $AB$ in midpoint $\implies BX || OO' \implies \angle ABX = 90^\circ.$

b) $OM = O'M, AM = MX \implies AOXO'$ is parallelogram $\implies AO = XO' = CO, OX = AO' = C'O'.$

Denote $\angle OAC = \angle OCA = \alpha, \angle O'AC' = \angle O'C'A = \beta,$ $\angle XO'A = \angle XOA = \varphi \implies$

$\angle CAC' = 180^\circ = \alpha + (180^\circ - \varphi) + \beta \implies \alpha + \beta = \varphi.$ $\angle COX = 360^\circ - (180^\circ - 2 \alpha) - \varphi = 180^\circ + \alpha - \beta.$ $\angle XO'C' = \varphi + \angle AO'C' = \varphi + (180^\circ - 2 \beta) = 180^\circ + \alpha - \beta.$ $\angle XO'C' = \angle COX \implies \triangle COX = \triangle XO'C' \implies CX = C'X.$

Corollary

Let points $D \in \omega, D' \in \Omega,$ and $A$ be collinear. Then $DX = D'X.$

Therefore $X$ is the crosspoint of the bisectors $CC'$ and $DD'.$

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Remarkable point for pair of similar triangles

Let $\triangle ACD \sim \triangle AC'D'.$

Let the points $O$ and $O'$ be the circumcenters of $\omega (\triangle ACD)$ and $\Omega (\triangle AC'D').$

Let point $M$ be the midpoint of $OO'.$

The point $X$ is symmetric to $A$ with respect point $M.$

Prove:

a) point $X$ be the crosspoint of the bisectors $CC'$ and $DD'.$

b) $\angle ABX = 90^\circ.$

Proof

$OM = MO', AM = MX \implies AOXO'$ is parallelogram $\implies$ $CO = AO = XO', C'O' = AO' = XO.$ Denote $\angle ADC = \angle AD'C' = \alpha \implies \overset{\Large\frown} {AC} = \overset{\Large\frown} {AC'} = 2\alpha \implies$

$\angle AOC = \angle AO'C' = 2 \alpha \implies$ $\angle COX = 360^\circ - \angle AOX - 2 \alpha = \angle C'O'X \implies$ $\triangle XOC = \triangle C'O'X \implies CX = C'X.$ $\angle CXC' = \angle CXO + \angle OXO' + \angle O'XC'$ $\angle CXC' = \angle CXO + (180^\circ - \angle AOX) + \angle OCX$ $\angle CXC' = (180^\circ - \angle AOX) + (180^\circ - \angle COX).$ $\angle CXC' = 360^\circ - \angle AOX - \angle COX) = \angle AOC = 2 \alpha.$ Similarly, $DX = D'X, \angle DXD' = 2 \angle ACD.$

The statement that $\angle ABX = 90^\circ$ was proved in the previous section.

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Remarkable point’s problems

Problem 1

Let a convex quadrilateral $ABCD$ be given, $E = AC \cap BD.$

Let $M$ and $N$ be the midpoints of $AC$ and $BD,$ respectively.

Circumcircles $ABE$ and $CDE$ intersect a second time at point $K.$

Prove that points $M, K, N,$ and $E$ are concyclic.

Problem 2

Let triangle $\triangle ABC$ be given.

Let point $D$ lies on sideline $BC.$

Denote the circumcircle of the $\triangle ABD$ as $\omega$ , the circumcircle of the $\triangle ACD$ as $\Omega$ .

Let $X$ be the circumcenter of $\Omega.$

Let circle $ACD$ cross sideline $AB$ at point $E.$

Let the circumcircle of the $\triangle BCE$ cross $\omega$ at point $F.$

Prove that $\angle BFX = 90^\circ.$

Problem 3

The circles $\omega$ and $\Omega$ are crossed at points $A$ and $B,$ points $C \in \Omega, D \in \omega.$

Let $AC$ be the tangent to $\omega, AD$ be the tangent to $\Omega.$

Point $E$ is symmetric $A$ with respect to $B.$

Prove that points $A, C, D,$ and $E$ are concyclic.

Problem 4

The circles $\omega$ and $\Omega$ are crossed at points $A$ and $B,$ points $C \in \Omega, D \in \omega.$

Let $AC$ be the tangent to $\omega, AD$ be the tangent to $\Omega.$

Points $E \in \omega$ and $F \in \Omega$ lye on bisector of the angle $\angle CAD.$

Points $E' \in \omega$ and $F' \in \Omega$ lye on external bisector of the angle $\angle CAD.$

Prove that $DE, CF,$ and bisector $E'F'$ are tangent to the circle $\theta = BEF,$

bisector $EF$ is tangent to the circle $\Theta =BE'F'.$

Solutions

Solutons are clear from diagrams.

In each case we use remarcable point $X,$ as the point of bisectors crossing.

Solution 1

We use bisectors $AC$ and $BD$ . $\angle XME = \angle XNE = \angle XKE = 90^\circ.$ The points $X, M, K, N,$ and $E$ are concyclic.

Solution 2 $AX = CX, DX = EX.$ We use bisectors of $AC$ and $DE.$ $\angle BFX = 90^\circ.$

Solution 3

We use bisectors of $AC$ and $AD$ .

$X$ is the circumcenter of $\triangle ACD, \angle ABX = 90^\circ.$

Circle $ACD$ is symmetric with respect diameter $BX.$

Point $E$ is symmetric to $A \in ACD$ with respect diameter $BX.$

Therefore $E \in ACD.$

Solution 4

Let $X = EE' \cap FF' \implies XE = XF, XE' = XF', XB \perp AB.$

Let $Y = AB \cap \theta, Z = AB \cap \Theta, M$ be midpoint $EF, M'$ be midpoint $E'F'$ .

$\angle EAE' = 90^\circ.$ We need prove that $X \in \theta$ and $X \in \Theta.$

Denote $\angle E'AD = \alpha \implies \angle EAD = \angle EAC = 90^\circ - \alpha \implies \angle CAF' = \alpha.$

The angle between a chord and a tangent is half the arc belonging to the chord.

$AD$ is tangent to $\Omega \implies \overset{\Large\frown} {AF'} = 2 \alpha \implies \angle ACF' = \alpha.$

$\angle AF'C = 180^\circ - 2\alpha, \angle CF'F = \angle CAF = 90^\circ - \alpha = \angle ACF \implies$

$FF'$ is diameter $\Omega.$ Similarly, $EE'$ is diameter $\omega.$ $\angle XEF = \angle E'EA = \alpha, \angle EFX = \angle AFF' = \alpha.$ $\angle EBX = 90^\circ - \angle EBA = 90^\circ - \angle EE'A =\alpha \implies X \in \theta \implies ZX$ is tangent to $\theta.$

Let $T$ be the spiral similarity $T(\omega) = \theta$ centered at $B.$

Points $A, F,$ and $E$ are collinear $\implies T(A) = F.$ Points $E', X,$ and $E$ are collinear $\implies T(E') = X.$

Therefore $T(D) = E \implies DE$ is tangent to $\theta.$

Similarly, $CF$ is tangent to $\theta.$

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Japan Mathematical Olympiad Finals 2018 Q2

Given a scalene $\triangle ABC,$ let $D$ and $E$ be points on lines $AB$ and $AC,$ respectively, so that $CA = CD, BA = BE.$

Let $\omega$ be the circumcircle of $\triangle ADE$ and $P$ the reflection of $A$ across $BC.$

Lines $PD$ and $PE$ meet $\omega$ again at $X$ and $Y,$ respectively.

Prove that $BX$ and $CY$ intersect on $\omega.$

Proof

Let $H$ be the orthocenter of $\triangle ABC.$

$AB = BE \implies$ Point $E$ is symmetrical to point $A$ with respect to height $BH.$

Point $D$ is symmetrical to point $A$ with respect to height $CH \implies$

$AH = EH = DH \implies \omega$ is centered at $H \implies$

$\omega$ is symmetrical with respect to heightline $AHP \implies$

$X$ is symmetrical to point $E$ with respect to height $AH,$

$Y$ is symmetrical to point $D$ with respect to height $AH \implies$

$\angle XAD = \angle EAY = \angle CAY.$

The isosceles triangles $\triangle ABE \sim ACD \implies$ $\frac {AB}{AC} = \frac {AE}{AD} = \frac {AX}{AY} \implies \triangle AXB \sim \triangle AYC \implies$