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MithsApprentice wrote: Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_{1}$ and $E_{1}$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_{2}$ and $E_{2}$ the points on sides $BC$ and $AC$, respectively, such that $CD_{2}=BD_{1}$ and $CE_{2}=AE_{1}$, and denote by $P$ the point of intersection of segments $AD_{2}$ and $BE_{2}$. Circle $\omega$ intersects segment $AD_{2}$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_{2}P$. SOLUTION Because of the homothety and the equality of segments (isotomic points and tangents), we get: $\frac{AQ}{QD_{2}}=\frac{AE_{1}}{E_{1}Z}=\frac{CE_{2}}{E_{1}C+CD_{2}}=\frac{CE_{2}}{CD_{1}+BD_{1}}=\frac{CE_{2}}{BC}$ $(1)$ Now we apply Menelaus's Theorem to the triangle $AD_{2}C$: $\frac{PD_{2}}{PA}=\frac{BD_{2}\cdot E_{2}C}{BC \cdot E_{2}A}=\frac{CE_{2}}{BC}\cdot \frac{BD_{2}}{E_{2}A}=\frac{CE_{2}}{BC}$ $(2)$ From $(1)$ and $(2) \Longrightarrow q.e.d.$ Nick Rapanos

I'm sorry for reviving an old thread, but I have an alternate solution (though not as short as the above one) to this problem. It suffices to prove that $\frac{AQ}{AD_2}=\frac{PD_2}{AD_2}$. We will find the values of these fractions separately, and prove that they are equal. Before we begin, for the sake of convenience, let us have $AE_1=a$, $BD_1=b$, and $D_1D_2=c$. First, we find the value of $\frac{AQ}{AD_2}.$ Since $BD_1=CD_2$, the $A$-excircle meets $BC$ at $D_2$, and so $Q$ is diametrically opposite to $D_1$. Hence, the line through $Q$ parallel to $BC$ is tangent to the incircle. Furthermore, let the homothety centered at $A$ that takes $Q$ to $D_2$ take $B$ to $T$. We have $\frac{AQ}{AD_2}=\frac{AB}{AT}=\frac{a(a+b)}{a^2+3ab+ac+2b^2+bc}$; the length of $AT$ can be found easily using the relation $\frac{a}{a+b}=\frac{a+2b+c}{AT}$. Now, we find the value of $\frac{PD_2}{AD_2}$. Assign a mass of $a, b, b+c$ to points $A, B, C$, respectively. Then $P$ has a mass of $a+2b+c$. This implies $\frac{AP}{PD_2}=\frac{2b+c}{a}$. Adding one to each sides, $\frac{AD_2}{PD_2}=\frac{a+2b+c}{a}$. Reciprocating both sides, we have $\frac{PD_2}{AD_2}=\frac{a}{a+2b+c}$. Simple algebra reveals that $\frac{a(a+b)}{a^2+3ab+ac+2b^2+bc}=\frac{a}{a+2b+c}$, and we have $AQ=D_2P$, as desired. $\blacksquare$

Quote: Let $ABC$ be a triangle and let $w(I)$ be its incircle. Denote by $D_{1}$ and $E_{1}$ the points where $\omega$ is tangent to sides $BC$ , $AC$ respectively. Denote by $D_{2}$ , $E_{2}$ the points on sides $BC$ and $AC$ respectively, such that $CD_{2}=BD_{1}$ , $CE_{2}=AE_{1}$ and denote by $P$ the point of intersection of segments $AD_{2}$ and $BE_{2}$ . Circle $w$ intersects segment $AD_{2}$ at two points, the closer of which to the vertex $A$ is denoted by $Q$ . Prove that $AQ=PD_{2}$ . Proof. Is well-known property $Q\in ID_1\ \ (*)$ . Therefore, $\frac {AQ}{AD_2}=\frac {h_a-2r}{h_a}=\frac {ah_a-2ar}{ah_a}=\frac {2pr-2ar}{2pr}=\frac {p-a}{a}\ \ (1)$ . Apply Menelaus' theorem to $\overline {BPE_2}/AD_2C\ \ :\ \ \frac {BD_2}{BC}\cdot\frac {E_2C}{E_2A}\cdot\frac {PA}{PD_2}=1$ $\implies$ $\frac {p-b}{a}\cdot\frac {p-a}{p-c}\cdot\frac {PA}{PD_2}=1$ $\implies$ $\frac {PA}{p}=\frac {PD_2}{p-a}=\frac {AD_2}{a}$ $\implies$ $\frac {PD_2}{AD_2}=\frac {p-a}{a}\ \ (2)$ . From the relations $(1)$ , $(2)$ obtain $AQ=PD_2$ . ================================== $(*)\ \blacktriangleright$ Denote $\{D_1,Q_1\}=w\cap D_1I$ .Prove easily that $\frac {AD}{Q_1D_1}=\frac {h_a}{2r}=\frac {ah_a}{2ar}=\frac {2pr}{2ar}=\frac pa$ and $\frac {DD_2}{D_1D_2}=\frac {p|b-c|}{a}\cdot \frac {1}{|b-c|}=\frac pa$ . In conclusion, $\frac {AD}{Q_1D_1}=\frac {DD_2}{D_1D_2}$ , i.e. $Q_1\in w\cap AD_2$ $\implies$ $Q_1\equiv Q$ .

In my solution, I used also the fact that $Q\in D_1I$ which is new for me, I just remarqued it from the diagram and proved it, and I think if we knew this result, the problem becomes easy.

Dear mathlinkers, this problem come from Lalesco a geometer of the beginning of the XX centuries... Why P which is the Nagel point is not used in order to have a synthetic proof? Sincerely Jean-Louis

jayme wrote: Dear mathlinkers, this problem come from Lalesco a geometer of the beginning of the XX centuries... Why P which is the Nagel point is not used in order to have a synthetic proof? Sincerely Jean-Louis I am trying to get a proof using the fact that $P$ is the nagel point but I am not able to complete it. May be someone can. From an homothety with centre A and ratio $\frac{s-a}{s},$ the point $D_2$ maps to $Q$ and hence $\frac{AD_2}{AQ}=\frac{s}{s-a}$ and hence $\frac{AP}{AQ}+\frac{D_2P}{AQ}=1+\frac{a}{s-a}$ It suffices to prove that $\frac{AP}{AQ}=\frac{a}{s-a}\Longleftrightarrow \frac{AP}{AD_2}=\frac{a}{s}$ and from here, I am not able to complete the proof. If someone knows the length of the nagel point from the vertex...

Nagel point $P$ of $\triangle ABC$ is the incenter of its antimedial triangle $\triangle A_0B_0C_0$ and orthogonal projections $M,M_0$ of $A,A_0$ on $BC$ are symmetric about the midpoint of $BC.$ Let the incircle $(P,2r)$ of $\triangle A_0B_0C_0$ touch its side $B_0C_0$ at $D_0$ and let $R \equiv AP \cap A_0M_0.$ Since $A_0D_1$ is the $A_0$- Nagel ray of $\triangle A_0B_0C_0,$ it follows that $A_0D_1 \parallel AR$ $\Longrightarrow$ quadrilateral $QRA_0D_1$ is a parallelogram, hence $A_0R=QD_1=PD_0=2r.$ $T$ is the orthogonal projection of $P$ onto $BC,$ since $D_0T=A_0M_0.$ Then $PT=RM_0$ $\Longrightarrow$ $\overline{D_2T}=\overline{D_2M_0}=\overline{D_1M}.$ If the projections of $\overline{PD_2},\overline{AQ}$ on $BC$ are equal, then $PD_2=AQ,$ as desired.

Here is a rather nice solution using barycentric coordinates: Let $A$ be $(1,0,0)$, $B$ be $(0,1,0)$, and $C$ be $(0,0,1)$. Let the side lengths of the triangle be $a,b,c$ and the semi-perimeter $s$. Now, $CD_1=s-c$, $BD_1=s-b$, $AE_1=s-a$, $CE_1=s-c$. Thus, $CD_2=s-b$, $BD_2=s-c$, $AE_2=s-c$, $CE_2=s-a$. Therefore, the non-normalized coordinates of $D_2$ are $(0,s-b,s-c)$ and of $E_2$, $(s-a,0,s-c)$. Clearly then, the non-normalized coordinates of $P$ are $(s-a,s-b,s-c)$. Normalizing, we have that $D_2$ is at $\left(0,\frac{s-b}{a},\frac{s-c}{a}\right)$, $E_2$ at $\left(\frac{s-a}{b},0,\frac{s-c}{b}\right)$, and $P$ is at $\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right)$ Now, we find the point $Q'$ inside the triangle on the line $AD_2$ such that $AQ'=D_2P$. It is then sufficient to show that this point lies on the incircle. $P$ is the fraction $\frac{s-a}{s}$ of the way "up" the line segment from $D_2$ to $A$. Thus, we are looking for the point that is $\frac{s-a}{s}$ of the way "down" the line segment from $A$ to $D_2$, or, the fraction $1-\frac{s-a}{s}$ of the way "up". Thus, $Q'$ has normalized $x$-coordinate $1-\frac{s-a}{s}=\frac{a}{s}$. As the line $AD_2$ has equation $(s-c)y=(s-b)z$, it can easily be found that $Q'$ lies at $\left(\frac{a^2}{as}, \frac{(s-a)(s-b)}{as}, \frac{(s-a)(s-c}{as}\right)$, or in non-normalized coordinates with sum $as$, at $(a^2,(s-a)(s-b),(s-a)(s-c)$. Recalling that the equation of the incircle is $a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0$, we must show that this equation is true for $Q'$'s values of $x,y,z$. Plugging in our values, this means showing that $a^2(s-a)^2(s-b)(s-c)+a^2(s-a)[b^2(s-c)+c^2(s-b)]+as[a^2(s-a)^2+(s-a)(s-b)^3+(s-a)(s-c)^3]=0$. Dividing by $a(s-a)$, this is just $a(s-a)(s-b)(s-c)+a[b^2(s-c)+c^2(s-b)]+a^2s(s-a)+s(s-b)^3+s(s-c)^3=0$. Plugging in the value of $s$: $\frac{a(-a+b+c)(a-b+c)(a+b-c)}{8}+\frac{ab^2(a+b-c)}{2}+\frac{ac^2(a-b+c)}{2}+\frac{a^2(a+b+c)(-a+b+c)}{4}+\frac{(a+b+c)(a-b+c)^3}{16}+\frac{(a+b+c)(a+b-c)^3}{16}=0$. $2a(-a+b+c)(a-b+c)(a+b-c)+8ab^2(a+b-c)+8ac^2(a-b+c)+4a^2(a+b+c)(-a+b+c)+(a+b+c)(a-b+c)^3+(a+b+c)(a+b-c)^3=0$ $2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0$ The first bracket is just: $-a^3-b^3-c^3+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b-2abc+4ab^2+4b^3-4b^2c+4ac^2-4bc^2+4c^3=-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc$ and the second bracket is $-2a^3+4a^2b+4a^2c+6ab^2+6ac^2-12abc$. Dividing everything by $2a$ gives $-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc+(a+b+c)(-a^2+2ab+2ac+3b^2+3c^2-6bc)$, which is $0$, as desired. As $Q'$ lies on the incircle and $AD_2$, $Q'=Q$, and our proof is complete.

Some days ago i saw another problem solved by using this Lemma but I couldn't solve.This one can be solved by following Lemma too: Lemma: $I$(incenter) , $G$(centroid),$N$(Nagel point) lie in that order on a line (Known as the Nagel line) & $GN=2IG$ Proof:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=518404 Draw a line through $D_{1}$ & $G$ so it intersects$PQ$ at$ F$.Name midpoint of $BC$ : $D"$ As $PI$ bisects $QD_{1}$ & $G$ is on $PI$ such that $GP=2IG$(according to theLemma) so $G$ is the centroid of $\vartriangle PQD_{1}$ so$F$ is the midpoint of $ QP$ (1) In $AD_{1}D_{2}$: $AD"$ bisects $D_{1}D_{2}$ and $G$ is on it such that $AG=AGD"$ so $G$ is the centroid of$\vartriangle AD_{1}D_{2}$ and $F$ is the midpoint of $AD_{2}$. (2) By using (1) & (2) we can conclude .

The above lemma is awesome. See here and here. The key point is that the Nagel point of a triangle (in this case, $P$) is the incenter of its anticomplementary triangle.

Let $\omega_a$ be the excircle of $\triangle{ABC}$ opposite to vertex $A.$ $\omega_a$ touches $BC$ at $D_2$ Let $\omega_a$ touches $AC$ at $H.$ Then $\angle{D_2HE}=\frac{\angle{C}}{2}$ It is well known lemma that $Q,I,D_1$ are collinear. $\angle{E_1D_1C}=\angle{D_1E_1C}=\pi-\frac{\angle{C}}{2}$ So, $\angle{AE_1Q}=\frac{\angle{C}}{2}=\angle{D_2HE}$ Hence $D_2H || QE_1$ So $\frac{AQ}{AD_2}=\frac{AE_1}{AH}=\frac{s-a}{s}$ Applying menelus theorem ($\triangle{AD_2C},\overline{E_2PB}$) it is easy to obtain $\frac{PD_2}{AD_2}=\frac{s-a}{s}$ Hence $AQ=D_2P$

Let $ M$ is the midpoint of $ BC \implies D_1M=D_2M $ , $D_1Q $ is the diameter and $ I $ is the center of $\omega$, implies $IM \parallel AD_2$ by using midpoint theorem in $ \triangle QD_1D_2 $ and $ 2IM=QD_2$ $ AD_2 \cap BE_2 =P$ (Nagel point ) $ AM \cap IP =G $(Centroid) and by using the fact that $ 2IG=GP$ we get that $AP=2IM=QD_2 \implies AQ=D_2P $

This statement $ AQ = D_2P$ can be used to prove the existence of nagel line and $ 2IG=GN$ see here:http://www.artofproblemsolving.com/community/c3141h1088525_nagel_line__aq__d_2p

Solution found with K6160 . For convenience, let us rephrase the problem: Let $\triangle ABC$ be a triangle, and let $D, A_1$ be the points of tangency of the incircle, $A$-excircle, respectively, with side $\overline{BC}.$ Let $D'$ be the intersection of $AA_1$ and the incircle that is closer to $A$, and let $N$ be the Nagel point of $\triangle ABC.$ Then $AD' = A_1N.$ Solution: Let $I, G$ be the incenter, centroid, respectively, of $\triangle ABC.$ We proceed with complex numbers: WLOG let $I$ be the origin, and assign complex numbers $a, b, c, d, d', a_1, n$ to $A, B, C, D, D', A_1, N$, respectively. It is well-known that $\overline{DD'}$ is a diameter of $\omega$, so $d' = -d.$ Now, in order to show that $AD' = A_1N$, it suffices to prove that $\overline{AA_1}$ and $\overline{D'N}$ share the same midpoint, i.e. $d' + n = a + a_1.$ This is equivalent to $n = a + a_1 + d$, but since $\overline{A_1D}$ and $\overline{BC}$ share the same midpoint, we have $a_1 + d = b + c.$ Therefore, it suffices to prove that $n = a + b + c = 3g.$ Because $I$ is the origin, this is equivalent to $IN = 3IG$, but this is just a well-known property of the Nagel line. $\square$

From the lemma we know that $D_1Q$ is a diameter of the incircle. Let the incenter of ABC be I, its centroid be G, and the midpoint of BC be M. Note that P is the Nagel point of ABC. From the previous problem, we know that the dilation centered at G with ratio −2 sends M to A and I to P, and hence sends segment IM to PA, thus PA = 2IM. On the other hand, a dilation centered at $D_1$ with ratio 2 sends IM to $QD_2$, so $QD_2$ = $2IM$ = $PA$. Therefore, AQ = PA−QP = $QD_2$ − $QP$ =$D_2P$. (Lemma used is the diameter of incircle)

Clearly, $P$ is the Nagel point of triangle $ABC$. Reflect $A$ about the incenter to $A'$. If we can show that $A'D_1PD_2$ is a parallelogram, we will be done, for reflecting back about $Q$ yields $A'D_1=AQ=PD_2$, as it is well known that $Q$ is the antipode of $D_1$ with respect to the incircle. We claim that triangles $AA'P$ and $ABC$ have concurrent centroids $G$. Indeed, by the Nagel line, $I$, $G$, and $P$ are collinear, but $I$ is the midpoint of $AA'$, so $G$ is indeed the centroid of triangle $AA'P$. Denoting $M$ as the midpoint of $BC$, it is now clear that $A', M$ and $P$ are collinear, whence $D_1A'D_2P$ is a parallelogram, so we are done.

WLOG let $b>c$. Let $D$ be the foot of the altitude from $A$ to $BC$ and let $D_3$ be the foot of the altitude from $P$ to $BC$. It suffices to show that $D_2D_1=D_3D=b-c$. By mass points we have $D_3D=D_2D\left(\frac{2a}{a+b+c}\right)$ and since $BD=\frac{c^2-b^2+a^2}{2a}$ we know that $D_2B=\frac{a+b-c}{2}$ we have that $D_2D=\frac{ab-ac-c^2+b^2}{2a}$ and $D_3D=b-c$, done.

rcorreaa wrote: We use Barycentric Coordinates with $ABC$ being the reference triangle. Since $P$ is the Nagel Point, $$P= (s-a : s-b : s-c)= (\frac{b+c-a}{a+b+c},\frac{c+a-b}{a+b+c},\frac{a+b-c}{a+b+c}) \qquad (\star)$$ Let $I$ the incenter of $ABC$. It's well known that $I$ is the midpoint of $D_1Q$ and $I=(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c})$. Hence, $$Q=(\frac{2a}{a+b+c},\frac{2b}{a+b+c}-\frac{s-c}{2a},\frac{2c}{a+b+c}-\frac{s-b}{2a})$$so $\frac{P+Q}{2}= (1,1-\frac{s-c}{2a},1-\frac{s-b}{2a})= \frac{A+D_2}{2}$, since $D_2=(0 : s-c : s-b)$. Thus, the midpoint of $PQ$ coincides with the midpoint of$AD_2$, so $AQ=D_2P$, as desired. $\blacksquare$ Very ingenious solution.

Note that $P$ is the Nagel point and let $I$ and $G$ be the incenter and centroid, respectively. Then, $Q$ is the antipode of $D_1$ with respect to $\omega.$ Also, $\triangle GAP\sim\triangle GMI$ since $AG/GM=PG/GI=2.$ Hence, $QD_2=2IM=AP.$ $\square$

This screams bary lmao. Note that $Q$ is the antipode of $D_1$ on the incircle. We consider bary with $\triangle ABC$ as the reference triangle. Then, $I = (a:b:c), D_1 = (0:s-c: s-b)$, and thus, $Q= 2I - D = (\frac{2a}{a+b+c}, \frac{2b}{a+b+c} - \frac{s-c}{a}, \frac{2c}{a+b+c} - \frac{s-b}{a})$. Then, note that $P = (s-a: s-b: s-c)$ is the Nagel point. Then, it suffices to show that $P+Q = A+D_2$, and since they're collinear we check the first coordinate, so \[\frac{-\frac12 a + \frac12 b + \frac12c}{\frac{a+b+c}{2}} + \frac{2a}{a+b+c} = 1 + 0\]which is true, so $AQ = D_2P$ and we're done. $\blacksquare$.

$AP=2IM=QD_2$, where $M$ is the midpoint of $BC$.

Note that $Q$ is the antipode of $D$ in the incircle. $I$ is the midpoint of $QD$ and $M$ the midpoint of $D_1D_2.$ Thus, $IM=\frac12 QD_2.$ Also, $I,G,P$ are collinear with $G$ one-third of the way from $I$ to $P$. Since $MI\parallel AP$, $\triangle MIG\sim \triangle APG\implies AP=2MI$ which implies result.

actually human solution: Claim: $Q$ is the reflection of $D_1$ over $I$ Consider the $A$-excircle, and let the tangents from $A$ to the $A$-excircle have length $x$, then considering the tangents from $B,C$ gives $x - b + x - c = a$, so $x = s$, then $D_2$ is the tangency of the $A$-excircle and $BC$ by length equivalence. Then by Fact 5, $A$, and the centers of the incircle and the $A$-excenter are collinear, so we consider the homothety mapping the incircle to the $A$-excircle, this also maps $Q$ to $D_2$, so we are done. Now, we prove the result that $\frac{AQ}{QD_2} = \frac{D_2P}{PA}$, which finishes. Notice that $\frac{D_2P}{PA} = \frac{CE_2}{E_2A} \frac{BD_2}{BC} = \frac{s - a}{a}$ by ratio lemma. Then we use ratio lemma again, get $\frac{AQ}{QD_2} = \frac{AB}{BD_2} \frac{\sin ABQ}{\sin QBD_2}$, then drop perpendiculars $D_3, Q_2$ from $D_1,Q$ to $AB$, then we have $\frac{AB}{BD_2} \frac{\sin ABQ}{\sin QBD_2} = \frac{c}{s - c} \frac{QQ_2}{QD_1} $, now notice (where undefined points are to be defined analgously) that $2F_1F_2 = D_1D_3 + QQ_2$, so we can evaluate $QQ_2 = 2r - (s-b)\sin B$, so our desired equality is $\frac{s-a}{a} = \frac{c}{s-c} \frac{2r - (s-b) \sin B}{2r}$, which we can reduce to $\frac{b^2 - (a-c)^2}{4ac} = 1 - \frac{\sin B}{2 \tan \frac B2}$, which reduces further to $\frac{b^2-a^2-c^2}{4ac} = \frac{1}{2} - \cos^2 \frac B2$, which we can reduce further to $\frac{b^2 - a^2 - c^2}{4ac} = \frac 12 - \frac{1 + \cos B}{2}$, which rearranges to the Law of Cosines, so because all steps are reversible, we are done.

Very easy especially for U2. By Menelaus on triangle $AD_2C$ with respect to $\overline{BPE_2}$, we obtain $\frac{AP}{PD_2} = \frac a{s-a}$. Also, $$\frac{D_2Q}{D_2A} = \frac{2r}{h_A} = \frac as,$$so the result follows.

$Q$ is the antipode of $D_1$ with respect to the incircle, so the tangent to the incircle at $Q$ is parallel to $\overline{BC}$. If the triangle $\triangle$ formed by it and $\overline{AB},\overline{AC}$ has similarity ratio $k$ from $[ABC]$, then we have $$k=\frac{AQ}{AD_2}=\sqrt{\frac{[\triangle]}{[ABC]}}=\sqrt{\frac{k(s-a)}{s}} \implies \frac{AQ}{AD_2}=\frac{s-a}{s}.$$On the other hand, in barycentric coordinates is clear that $D_2=(0:s-b:s-c)=(0,\tfrac{s-b}{a},\tfrac{s-c}{a})$ and $P=(s-a:s-b:s-c)=(\tfrac{s-a}{s},\tfrac{s-b}{s},\tfrac{s-c}{a})$, hence $\frac{PD_2}{AD_2}=\frac{s-a}{s}$ as well, so we're done. $\blacksquare$

Note that $D_2$, $E_2$ are the respective extouch points. By Menelaus on $\triangle AD_2C$, we find that (in non-directed lengths) $$\frac{D_2P}{PA} = \frac{E_2C}{AE_2} \cdot \frac{BD_2}{CB} = \frac{s - a}{s - c} \cdot \frac{s - c}{a} = \frac{s - a}{a}.$$Also, by a homothety at $A$, we find that $Q$ is the antipode of $D_1$ wrt the incircle of $\triangle ABC$. Therefore if $E_3$ is the projection of the $A$-excenter of $\triangle ABC$ onto $\overline{AC}$, then $$\frac{AQ}{QD_2} = \frac{AE_1}{E_1E_3} = \frac{s - a}{a} = \frac{D_2P}{PA},$$which implies the conclusion.

Oh wow did I overkill this. Denote $BC, CA, AB, $ the semiperimeter, inradius, and $A-$exradius of $\Delta ABC$ by $a, b, c, s, r, R$ respectively. Observe that $D_2, E_2$ are the $A, B$ extouch points respectively (i.e. where the $A-, B-$ excentres are tangent to $BC, CA$). Introduce the $F-$intouch and extouch points respectively to be $F_1, F_2$. Now observe that $AD_2, BE_2, CF_2$ concur due to Ceva's theorem and the existence of the incentre. Therefore, from Van Aubel's theorem and some well-known lengths, we get $$\frac{AP}{PD_2} = \frac{AE_2}{E_2C} + \frac{AF_2}{F_2B} = \frac{CE_1}{E_1A} + \frac{BF_1}{F_1A} = \frac{(s-b)+(s-c)}{(s-a)}= \frac{a}{s-a},$$so $$\frac{PD_2}{AD_2} = \frac{s-a}{s}.$$Also, it is well known that the homothety at $A$ taking the incircle to the $A-$excircle also takes $Q$ to $D_2$, so we get $$\frac{AQ}{AD_2} = \frac{r}{R} = \frac{s-a}{s},$$where the final part follows by the exradius length formula given in EGMO Chapter 2. Therefore, we get $$\frac{PD_2}{AD_2} = \frac{s-a}{s} = \frac{AQ}{AD_2},$$and the result follows. $\square$

We set $A=(1,0,0)$ and etc. Now, we know that $D_1=(0:a+b-c:a-b+c)$ and $E_1=(a+b-c:0:-a+b+c)$. Thus, we also immediately obtain that $D_2=(0:a-b+c:a+b-c)$ and $E_2=(-a+b+c:0:a+b-c)$. Now, it is clear that $P=(-a+b+c:a-b+c:a+b-c)$ since it is the intersection point of the cevians $AD_2$ and $BE_2$. We now do some synthetic work. Claim : The incenter $I$ of $\triangle ABC$ is in fact the midpoint of $QD_1$. Proof : Consider the homothety centered at $A$ mapping the incircle to the $A-$excircle. This clearly maps $D_2 \to Q$. Since $BC$ is tangent to the $A-$excircle at $D_2$, it also follows that the line parallel to , $BC$ through $Q$ is tangent to the incircle. Thus, $Q$ is the diametrically opposite point of $D_1$ from which the claim clearly follows. We know that $I=(a:b:c)$. Further, since $Q$ lies on $AD_2$, $Q=(t:a-b+c:a+b-c)$. Thus, $[BQC]=2[BIC]$ which means, \begin{align*} \frac{t}{t+(a-b+c)+(a+b-c)} &= 2\frac{a}{a+b+c}\\ \frac{t}{t+2a} &= 2\frac{a}{a+b+c}\\ ta + t(b+c) &= 2at +4a^2\\ t &= \frac{4a^2}{b+c-a} \end{align*}Thus, we conclude that $Q=(4a^2: c^2 - (a-b)^2 : b^2 - (c-a)^2)$. Now, to show that $AQ = PD_2$, it clearly suffices to show that $[ABQ] = [PBD_2]$. Now, note that \[\begin{vmatrix} 0 &1 & 0 \\ 1 & 0 & 0\\ \frac{2a^2}{a^2+ab+ac} & \frac{c^2 - (a-b)^2}{2(a^2+ab+ac)} & \frac{b^2 - (c-a)^2}{2(a^2+ab+ac)} \end{vmatrix} = \frac{b^2 - (c-a)^2}{2a^2+ab+ac} = \frac{b^2 - (c-a)^2}{2a(a+b+c)}\]and \[\begin{vmatrix} 0 &1 & 0\\ 0 & \frac{a-b+c}{2a} & \frac{a+b-c}{2a}\\ \frac{-a+b+c}{a+b+c} & \frac{a-b+c}{a+b+c} & \frac{a+b-c}{a+b+c} \end{vmatrix} = \frac{(a+b-c)(-a+b+c)}{2a(a+b+c)} = \frac{b^2- (c-a)^2}{2a(a+b+c)}\]so we are done.

Has no one lenght bashed this? Denote by $I$ the incenter of $\triangle ABC$ and $M=QE_1 \cap BC$. Also let $p$ denote the semiperimeter of the triangle. Menelaus in $\triangle ACD_2$ gives: $$ \frac{AE_2}{CE_2} \cdot \frac{BC}{BD_2} \cdot \frac{D_2P}{AP}=1$$$$ \Rightarrow \frac{p-c}{p-a} \cdot \frac{a}{p-a} \cdot \frac{D_2P}{AP} =1 (1) $$ It is well-known that $Q$ is the antipode of $D_1$ in the incircle. In $\triangle QD_1M$, $D_1M=2r\cot(\frac{\angle C}{2})$, therefore $CM=2r\cot(\frac{\angle C}{2}) - (p-c)$ and $D_2M= 2r\cot(\frac{\angle C}{2}) - (p-c)+(p-b)$. Applying Menelaus in $\triangle ACD_2$, we get: $$\frac{AQ}{QD_2} \cdot \frac{D_2M}{CM} \cdot \frac{CE_1}{AE_1} = 1 (2)$$From $(1)$ and $(2)$ it suffices to prove: $$1+\frac{p-b}{2r\cot(\frac{\angle C}{2})-(p-c)}=\frac{a}{p-c}$$$$\Rightarrow \frac{2r\cot(\frac{\angle C}{2})}{p-b}=\frac{p-c}{p-b} + \frac{p-c}{a+c-p} (3)$$ Using Heron’s formula and the metric formulas for $\sin\frac{\angle C}{2}$ and $\cos\frac{\angle C}{2}$ we can compute: $2r\cot(\frac{\angle C}{2})=2(p-c)$. Plugging this in $(3)$ and doing the calculations we get that the problem is equivalent to: $$\frac{1}{p-b} = \frac{1}{a+c-p}$$which is true, as desired. $\blacksquare$