Let the incircle touch sides $AC,AB$ at $E,F$, and let $X=EF \cap BC$. Then $X$ lies on the polar of $A$ with respect to the incircle, so $A$ lies on the polar of $X$ and thus $AD$ is the polar of $X$. In particular, $IX \perp AD$, so $\angle XPD=90^{\circ}$. Since $AD,BE,CF$ are concurrent, $(XD;BC)=-1$, and it is known the last two equalities imply $\angle BPD=\angle DPC$.

As in the above solution, we want to show that the point of intersection of EF and BC lies on IX, with all points defined as above. In other words, we want to show that IX, EF, BC concur. We do this via the radical axis theorem. Consider the incircle (passing through D,E,F), the circle with diameter AI (passing through through E,F and X) and the circle with diameter ID (which passes through X). It is clear that the pair wise radical axes of these circles are EF, IX and BC. (The last one is because BC is tangent to the circle with diameter AI and the circle with diameter ID). Hence by the radical axis theorem, EF, IX and BC concur and we can finish as in the first solution.

Let $\overline{AD}$ meet the incircle at $M$ and let ray $IP$ meet line $BC$ at $K$. Put tangency points $E$, $F$ on $\overline{AC}$, $\overline{AB}$ and let $X = \overline{EF} \cap \overline{MD}$. Because $\overline{IK} \perp \overline{MD}$, we see that $\overline{KM}$ is a tangent to the incircle. Now $FMED$ is harmonic, so $(K,X;F,E)=-1 \implies (K,D;B,C) = -1$. Since $\angle KPD = 90^\circ$, we deduce $\angle BPD = \angle DPC$.

Dear Mathlinkers, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=469215 Sincerely Jean-Louis

In the triangle $ABC$ let $D,E,F$ be the points of contact of the incircle with the sides $BC,CA,AB$. Let $BC\cap EF=X$ By ceva's and menalaus, we have $(C,D;B,X)$ harmonic. Let $XI\cap AD=P'$. Let $\angle BAD=\theta _1$, $\angle BDA=\theta _2$ and $\angle P'XD=\theta$ We will prove $\theta+\theta _2=90^{\circ}$ Let $AF=AE=x,EC=DC=z,DB=FB=y$ $ID=r=\frac{\Delta}{s}$. After some simplification, we get $r=\sqrt{\frac{xyz}{x+y+z}}$ From Menalaus we have $\frac{CX}{BX}\cdot \frac{BF}{FA}\cdot \frac{AE}{EC}=1$ $\implies \frac{CX}{BX}=\frac{z}{y}\implies \frac{CB}{BX}=\frac{z-y}{y}\implies BX=\frac{(y+z)y}{z-y}$ $\therefore XD=XB+BD=\frac{y^2+yz}{z-y}+y=\frac{2zy}{z-y}$. So,${\tan \theta=\frac{ID}{XD}=\frac{r}{\frac{2zy}{z-y}}=\frac{\sqrt{\frac{xyz}{x+y+z}}}{\frac{2zy}{z-y}}}\implies \tan \theta =\frac{x(z-y)}{2\sqrt{xyz(x+y+z)}}$. In $\Delta ABD$ by applying sine rule $\frac{\sin \theta _1}{BD}=\frac{\sin \theta _2}{AB}$ from here we get $\cot \theta _2=\frac{\frac{y}{x+y}-\cos B}{\sin B}$ Substituting the values $\sin B=\frac{2\sqrt{xyz(x+y+z)}}{(x+y)(y+z)}$ and $\cos B=1-\frac{2zx}{(x+y)(y+z)}$ we get $\cot \theta_2=\frac{x(z-y)}{2\sqrt{xyz(x+y+z)}}$ $\implies \tan\theta =\cot \theta _2\implies \theta+\theta_2 =90^{\circ}\implies IX \perp AD\implies P'=P$ So, $(C,D;B,X)$ is a harmonic division and $DP\perp IX$, so $\angle BPD=\angle DPC$.

Let $IP\cap BC=X$ we need $(B,C;D,X)=-1$. $XD$ is tangent to the incircle and $XI\perp AD$ so $A$ lies on the polar of $X$ wrt the incircle so $X$ lies on the polar of $A$, which is $EF$ where $E$,$F$ are points incircle is tangent to $AC$ and $AB$. Since $AD$,$BE$,$CF$ concur it follows $(B,C;D,X)=-1$.

Let $E,F$ be tangency points of incircle of $ABC$ with $AC,AD$. $AFIPE$ is concyclic with diameter $AI$.Call it $ \omega$. Also note that circumcircle $IPD$ is tangent to $BC$ at $D$. Let $FE \cap BC=X $. Power of $X$ with respect to $\omega$ = $XE*XF$ = $XD^2$ = Power of $X$ with respect to circumcircle $IPD$. So , $X$ lies on the radical axis of these two circles $\implies X$ lies on $IP$. Also , $ P(B,C;D,X)$ is a harmonic pencil. But , $ X$ lies on $IP$. $ \implies XP \perp PD$. (As $IP \perp PD$.) This finally forces that $PD$ bisects $ \angle BPC $ internally.

Projective quickies... Here's my solution: Let $EF$ meet the line $BC$ again at $T$ where $E,F$ are the touch points of the in-circle with $CA,AB$ respectively. Let $\omega$ denote the in-circle. Then, let $AD$ meet $\omega$ again at $X$. Now, since, $T$ lies on the polar of $A$ w.r.t $\omega$ we conclude that by La Hire's theorem $A$ lies on the polar of $T$ w.r.t $\omega $ and so $TI$ is perpendicular to $AD$. So, $T,P,I$ are col-linear and $(B,C;D,T)=-1$ so we get that since, $PD$ and $PT$ are perpendicular $PD$ bisects $\angle BPC$.

Let $X=PI \cap BC$ in the projective plane and $E,F$ be the tangency points of the incircle $\omega$ with $CA$ and $AB$. Then $X$ lies on the polar of $D$ wrt $\omega$, so by La Hire $D$ lies on the polar of $X$ wrt $\omega$. But $AP \perp XI$ and $D$ lies on $AP$, so $AP$ is the polar of $X$ wrt $\omega$. As $AP \cap EF$ lies on $AP$, the polar of $X$ wrt $\omega$, by a well-known lemma we have $(X,AP \cap EF;E,F)=-1$ and therefore \[ -1 = (X,AP \cap EF;E,F) \overset{A} = (X,D;C,B). \]By another well known projective lemma, we get $\angle BPD = \angle DPC$ due to $\angle DIX = 90^{\circ}$.

[asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5, xmax = 33.4, ymin = -10.76, ymax = 6.74; /* image dimensions */pen zzttqq = rgb(0.6,0.2,0); draw((5.56,3.16)--(4.22,-6.02)--(18.58,-6)--cycle, linewidth(2) + zzttqq); /* draw figures */draw((5.56,3.16)--(4.22,-6.02), linewidth(2) + zzttqq); draw((4.22,-6.02)--(18.58,-6), linewidth(2) + zzttqq); draw((18.58,-6)--(5.56,3.16), linewidth(2) + zzttqq); draw(circle((8.074325126250272,-2.682749520331696), 3.331879107279445), linewidth(2)); draw((5.56,3.16)--(8.078965527978559,-6.014625396200585), linewidth(2)); draw((-4.122994986121502,-6.031619770175654)--(6.376979626391591,0.1843808638041138), linewidth(2)); draw((-4.122994986121502,-6.031619770175654)--(9.991488504436088,0.042301482286132064), linewidth(2)); draw((-4.122994986121502,-6.031619770175654)--(18.58,-6), linewidth(2)); draw((-4.122994986121502,-6.031619770175654)--(8.074325126250274,-2.6827495203316962), linewidth(2)); draw((4.22,-6.02)--(7.227972577185077,-2.9151222661982366), linewidth(2)); draw((7.227972577185077,-2.9151222661982366)--(18.58,-6), linewidth(2)); /* dots and labels */dot((5.56,3.16),dotstyle); label("$A$", (5.64,3.36), NE * labelscalefactor); dot((4.22,-6.02),dotstyle); label("$B$", (4.3,-5.82), NW * labelscalefactor); dot((18.58,-6),dotstyle); label("$C$", (18.66,-5.8), NE * labelscalefactor); dot((8.078965527978559,-6.014625396200585),linewidth(4pt) + dotstyle); label("$D$", (8.16,-5.86), NE * labelscalefactor); dot((8.074325126250274,-2.6827495203316962),linewidth(4pt) + dotstyle); label("$I$", (8.16,-2.52), NE * labelscalefactor); dot((4.777384995552246,-2.2014968215152084),linewidth(4pt) + dotstyle); label("$F$", (4.86,-2.04), NE * labelscalefactor); dot((9.991488504436088,0.042301482286132064),linewidth(4pt) + dotstyle); label("$E$", (10.08,0.2), NE * labelscalefactor); dot((-4.122994986121502,-6.031619770175654),linewidth(4pt) + dotstyle); label("$K$", (-4.04,-5.88), NW * labelscalefactor); dot((6.376979626391591,0.1843808638041138),linewidth(4pt) + dotstyle); label("$J$", (6.46,0.34), NE * labelscalefactor); dot((7.227972577185077,-2.9151222661982366),linewidth(4pt) + dotstyle); label("$P$", (7.3,-2.76), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Define $E$ to be the intersection of the incircle with $CA$ and $F$ the intersection of the incircle with $AB$. Let $K=EF\cap BC$ and $J$ to be the second intersection of $AD$ with the incircle. Notice that because $JEDF$ is a harmonic quadrilateral, $KD$ and $KJ$ are the tangents to the incircle. From this we have $KI\perp AD$, so $K, P, I$ are collinear. It's well-known that $(K, D; B, C)=-1$; combining this with $\angle{KPD}=90^{\circ}$ implies (by another well-known lemma) that $PD$ bisects $\angle{BPC}$, as desired.

Let the second intersection of $AD$ with the incircle be $K$, and let the incircle touch $AB$ and $AC$ at $E$ and $F$ respectively. By symmetry, we have $\text{ Tangent to the incircle at K}$, $BC$ and $IP$ concurrent at (say) the point $Q$. Now, $$ -1 = (K, D; E, F) \overset{K}{=} (Q, KD \cap EF; E, F) \overset{A}{=} (Q, B; D, C)$$$$\Rightarrow \angle BPD = \angle DPC$$, hence QED. EDIT: Well there are some minor mistakes and the solution need clarification. I will correct it later.

Dear Mathlinkers also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=486932 Sincerely Jean-Louis

Let $IP\cap BC=Q$, it suffices to show that $(B,C;D,Q)=-1$. Let $E,F$ be feet of $I$ on $AC,AB$, then it is equivalent to show that $Q,E$ and $F$ are collinear; as $(B,C;D,EF\cap BC)=-1$. Since $QI \perp AD$, and $D$ lies on polar of $Q$, $AD$ is the polar of $Q$, which implies that $Q$ lies on polar of $A$. Since polar of $A$ is $EF$, $E,F$ and $Q$ are collinear, as needed.$\square$

Let the in-circle touch $AB,AC$ at $E,F$ respectively. Let $EF \cap BC=X$. Then $(X,D;B,C)=-1$. Then it suffice to prove that $X,P,I$ collinear $\Leftrightarrow$ $(A,F,I,P,E)$ all lie in a circle. Invert wrt to incircle then $\triangle DEF$ is fixed. ${A,B,C}$ swaps to the midpoints of $EF,DF,DE$ respectively. Let it be ${A',B',C'}$. Then $$X' \mapsto (IB'C') \cap (IEF)$$and $$P' \mapsto IX' \cap (IA'D)$$. Let $P''$ be $EF \cap X'I'$.I claim $P'' \equiv P'$. Since $X'$ is the midpoint of the $D$ symmedian in $\triangle DEF$, then $P''$ is just the intersection of $DD$ and $EF$. So $P''D^2=P''A'^2=P''X'.P''I \implies P'',D,I,A'$ lie on a circle. So $P'' \equiv P'$. Hence $I,P,E,F,A$ all lie on circle and we are done.

ELMO Shortlist 2012 G3 wrote: $ABC$ is a triangle with incenter $I$. The foot of the perpendicular from $I$ to $BC$ is $D$, and the foot of the perpendicular from $I$ to $AD$ is $P$. Prove that $\angle BPD = \angle DPC$. Alex Zhu. Set : $K = EF \cap BC, AD \cap \odot(DEF) = X$ . $\angle BPD = \theta_1$ and $\angle CPD = \theta_2$ Claim : $AD$ is the Polar of $K$ w.r.t the incircle . (Proof) Due to Cevians we get that $-1 = (K,D;B,C) \stackrel{A} = (K,AD \cap EF;F,E)$ . See that $-1 = (X,D;F,E) \stackrel{X} = (XX \cap EF,AD \cap EF;F,E) = -1 \Longrightarrow KX$ is tangent to $\odot(DEF)$ Main Problem : $-1 = \frac{BK}{CK} \cdot \frac{CD}{BD} = - \frac{\sin \theta_1 \cos \theta_2}{\cos \theta_1 \sin \theta_2} = - \frac{\tan \theta_1}{\tan \theta_2} \Longrightarrow \theta_1 = \theta_2$. Thus, we obtain the desired conclusion that $\angle BPD = \angle CPD$

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); /* initialization */ pair A = (-6, 5); pair B = (-9, -3.5); pair C = (6, -3.5); pair I = incenter(A,B,C); pair D = foot(I, B, C); pair F = foot(I,A,B); pair E = foot(I,C,A); pair K = extension(E,F,B,C); pair L = -D+2*foot(I, A, D); pair P = extension(I,K,A,D); /* Draw objects */ draw(A--B--C--cycle, green+linewidth(1)); draw(incircle(A,B,C),darkgreen+linewidth(1)); draw(K--F--E,blue); draw(K--B,green); draw(A--D,red); draw(I--P--K,cyan); draw(P--B,darkblue+dotted); draw(P--C,darkblue+dotted); draw(K--L,darkblue+dotted); /* name objects*/ dot(A); label("$A$", A, dir(90)); dot(B); label("$B$", B, dir(200)); dot(C); label("$C$", C, dir(340)); dot(I); label("$I$" , I,dir(90)); dot(D); label("$D$" , D,dir(180)); dot(E); label("$E$" , E,dir(180)); dot(F); label("$F$" , F,dir(180)); dot(K); label("$K$" , K,dir(180)); dot(L); label("$L$" , L,dir(180)); dot(P); label("$P$" , P,dir(180)); [/asy][/asy] Let $\overline{EF} \cap \overline{BC}=K$.Also $K$ is the pole of $\overline{AD}$ wrt incircle by La-Hire.So $K,P,I$ are collinear and $\angle KPD = 90^{\circ}$.Also we have $(BC;KD)=-1$.Whence $\angle BPD = \angle DPC$ and we are done.

Well...this is an overused configuration: https://artofproblemsolving.com/community/c16_national_and_regional_contests https://artofproblemsolving.com/community/c6h1796484

Let $E$ and $F$ be the feet of the perpendicular from $I$ to $AC$ and $AB$, respectively. Define $K$ to be the intersection of $EF$ and $BC$, and define $P'$ to be the intersection of $IK$ and $AD$. Claim: $\angle KP'D = 90$. Proof: Let $T = AI \cap (DEF)$. Notice that $DTEF$ is harmonic because $AF$ and $AE$ are tangents. Since $KD$ is a tangent, $KT$ is also a tangent, so $\angle KTI = 90$. Now $\angle KP'D = 90$, as desired. This suffices to prove that $P = P'$. Therefore, we know that $(B,C;K,D) = -1$. Moreover, since $\angle KPD = 90$, $\angle BPD = \angle CPD$, as desired.

Let $E$ and $F$ be the other tangency points and let $AD$ meet the incircle again at $R$. Then, we know $REDF$ is harmonic, so the tangents at $R$ and $D$ and $EF$ are concurrent at a point $Q$. Then, clearly $PI$ also goes through $Q$ because $P$ is the midpoint of $RD$. We also know $(QD;BC) = -1$ since $AD, BE,$ and $CF$ are concurrent at the Gergonne point. This implies that $\angle BPD = \angle CPD$ from a well known lemma.

Let $\overline{EF} \cap\overline{BC} = K$. Then, it suffices to prove (by Right Angles and Bisectors) that $(K, D; B, C) = -1$. But, note that $K$ lies on $\overline{EF},$ so $K$ lies on the polar of $A$ wrt incircle (which is $\overline{AD}$, since $\overline{AD}$ is tangent to the incircle), so by La Hire's Theorem, $A$ lies on the polar of $K$ wrt incircle, which means that $(K, D; B, C) = -1,$ as desired.

Notice that $AFPIE$ is cyclic, so $\angle APE=\angle APF$. Let $\overline{BE}$ and $\overline{CF}$ intersect at $G$, which lies on $\overline{AD}$. Notice that $(\overline{PA},\overline{PG})$ and $(\overline{PE},\overline{PF})$ are symmetric about $\overline{AD}$, so by DDIT on complete quadrilateral $BFAECG$, $(\overline{PB},\overline{PC})$ is symmetric about $\overline{AD}$, as desired. $\square$

Let $E$ and $F$ be the projections of $I$ onto $AC$ and $AB$. Then by Ceva's Theorem, we see that $AD$, $CF$, and $BE$ concur, and from Ceva-Menelaus we find that $(\overline{EF} \cap \overline{BC} = X, D; B, C) = -1$. By EGMO $9.40$ we find that $XI \perp AD$, so $\angle XPD = 90^{\circ}$, so we finish by the right angles and bisector lemma.

Without loss of generality treat $AB \leq AC$. Let $F$ and $E$ be the tangency points of $AC$ and $AB$ with the incircle. Then $AEPIF$ is cyclic with diameter $AI$ and the circle with diameter $ID$ is tangent to the incircle (as the center of the former lies on $ID$). Hence the radical axis $BC$, $IP$ and $EF$ of these circles concur at $T$, say. On the other hand, the lines $AD$, $BE$ and $CF$ are concurrent by Ceva's theorem and the equal tangents to the incircle. Hence by the complete quadrilateral $BCEF$ with $BC \cap = EF$ and $BF \cap CE = A$ we obtain $(B, C; D, T) = -1$. Finally, since $\angle KPD = 90^\circ$, by the angle bisector harmonic property we conclude $\angle BPD = \angle CPD$.

Let's call incircle with $\omega$ Let $\omega$ tangents to $AC,AB$ at $E,F$ Let $EF \cap BC$ at $T$ With easy harmonic lemma we know that, $AD \cap \omega$ at $W$ then $TW,TD$ tangent to $\omega$ Thus, $T$ is pole of $WD$ so $TI$ is perpendicular to $AD$ at $P$ We have harmonic lemma with bisector and perpendicular which give us to finish problem with $(B,C;D,T)=-1$. Which is correct with menelaus and ceva .

Let $PI \cap BC = X$. Now if $(X,D;B,C) = -1$ the condition will be satisfied, by the configuration with right angles and bisectors. Let E and F be the the tangency points of the incircle on AC and AB. Now let $EF \cap BC = T$ and by the Ceva configuration we have that $(T,D;B,C) = -1$ $\Rightarrow$ if we wanted to show that $(X,D;B,C) = -1 = (T,D;B,C)$ we only need to prove that $T \equiv X$, which will be true if $X \in FE$. So now we want to show that $X \in FE$. We know that XD is tangent to the incircle in D and $DP \perp XI$, where I is the incircle $\Rightarrow$ DP is the polar of X. We know that $A \in PD$, by the statement of the problem $\Rightarrow$ A lies on the polar of X. Now by La Hire we get that since A lies on the polar of X we must have that X lies on the polar of A. But since AE and AF are the tangent lines from A to the incircle it follows that the polar of A coincides with EF $\Rightarrow$ $X \in EF$, which is exactly what we wanted to prove so we are ready.

Let $E$ and $F$ be the feet of the perpendiculars from $I$ to $AC$ and $AB$, respectively. We claim that $EF$, $BC$, and $IP$ concur at a point $X$. We start by letting $X=EF\cap BC$. Since $EF$ and $BC$ are the polars of $A$ and $D$, respectively, with respect to the incircle $\Gamma$, $X$ is the pole of line $AD$, so $XI \perp AD$ and thus $X$ lies on $IP$ as well. (This can also be proven with radical center on $(AEIF)$, $(IPD)$, and $\Gamma$.) Let $Y=EF\cap AD$ and $Z=\Gamma \cap AD$, $Z\neq D$. Then we have: \[-1=(DZ;FE) \overset{D}{=} (XY;FE) \overset{A}{=} (XD;BC).\] Since we also have $\angle XPD = 90^{\circ}$, by the Right Angles and Bisectors configuration we have $\angle BPD = \angle DPC$, and we are done.

Let $\triangle DEF$ be the intouch triangle of $\triangle ABC$. Let $T=\overline{BC} \cap \overline{IP}$. Let $X \neq D$ be the point on the incircle such that $\overline{TX}$ is tangent to the incircle. Note that $P$ is the pole of $T$ with respect to the incircle, so $\overline{AXPD}$ is the polar of $T$. Due to this collinearity $-1=(XD; FE)$ which implies $F$, $E$, and $T$ are collinear. Hence $-1=(XD;FE) \stackrel{(DEF)}= (TD; BC)$ so by right angles and bisectors $\angle BPD = \angle DPC$.

Nice projective quickie. My solution: Let the incircle be tangent to $AB,AC$ at $F,E$ respectively. Then $AFPIE$ is cyclic. And $(IPD)$ is tangent to BC. by radical axis on incircle $(PID)$ and $(AI)$ we get that $BC,FE,IP$ concur, say at G. Then we have \begin{center} $(GD;BC)=-1$ \end{center} and $\angle GPD = 90^{\circ}$. And hence $PD$ bisects $\angle BPC.$ $QED$

Denote the tangency points of incircle with $AC$ and $AB$ to be $E$ and $F$ respectively. Let $X=PI \cap BC.$ By radical axis theorem, $FE, PI, BC$ concur at $X$ (this is because $(PID)$ is tangent to incircle and $IPFAE$ is cyclic since $\angle API=\angle AEI=90$). By Ceva, we know $AD, BE, CF$ concur, so by Menelaus we get $(X, D ; B, C)=-1.$ Since $\angle XPD= 90,$ we know $PD$ bisects $\angle BPC$ (follows immediately from law of sines).

Lines $\overline{AD}$, $\overline{BE}$, $\overline{CF}$ are known to be concurrent, and hence extending $\overline{EF}$ to intersect $\overline{BC}$ at $X$ gives $(X,D;B,C) = -1$ due to EGMO 9.11. Then, because of EGMO 9.18, it suffices to prove that $\overline{IX} \perp \overline{AD}$. However, we can simply note that $X$ is the polar of $\overline{AD}$ with respect to the incircle, which implies the desired result. $\square$

Kind of an overexplained solution to an otherwise easy problem. Let the incircle be $\omega$. $E= \omega \cap AC$,$F=\omega \cap AB$ and $T=EF \cap BC$. By Ceva-Menelaus,if $X=EF \cap AD$:- $(BC;DT) \stackrel{A}{=} (EF;XT)=-1$ implying $AD$ is the polar of $T$ with respect to $\omega$ since the polar is the locus points dividing the chords passing through $T$ harmonically with respect to $T$. Note that this implies if $AD \cap \omega =K \neq D$, then $TD,TK$ are tangents to $\omega$. Since $P$ is obviously the midpoint of $DK$, it follows that $\overline{I-P-T}$ are collinear. Now the required follows from the right angle-angle bisector-harmonic lemma from the book $EGMO$.

Invert at $I$. The problem becomes: Quote: Let $DEF$ be a triangle with circumcenter $I$ and let $A,B,C$ be the midpoints of $EF,FD,DE$. Let $P$ lie on $EF$ such that $DP$ is tangent to the circumcircle. Prove $\measuredangle PBI=\measuredangle ICP$. Invert at $D$. The problem becomes: Quote: Let $DEF$ be a triangle. Let $B,C,I$ be the reflections of $D$ over $F,E,EF$. Let $P$ lie on the circumcircle with $DP\parallel EF$. Prove $\measuredangle PBC=\measuredangle BCP$. This is easy: $DP,EF$ share a perpendicular bisector so homothety at $D$ with scale factor $2$ finishes.

Two lemmas destroy this problem 1) InCiRcLe PoLaRs LeMmA 2) BuY TwO GeT OnE FrEe LeMmA