Very fast complex bash: take $\Gamma$ to be the unit circle so that $$|a|=|b|=|c|=1$$$$d=\frac{a^2+ab+ac-bc}{2a}$$$$e = b + c - d = \frac{ab+ac+bc-a^2}{2a}$$$$p = -a$$$$\frac{e-p}{b-c} = \frac{(a+b)(a+c)}{2a(b-c)} \in i\mathbb{R}$$$\blacksquare$

Easy problem! In ∆PAC, C is right angle,PC=2RcosB. Also,EC=BD=2RsinCcosB. If E' is the foot of perpendicular from P to CB=>CE=CE'=>E and E'coincides.

Reflexions of the Orthocenter

[asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -12.711985344640167, xmax = 19.07233877485888, ymin = -8.194332200614756, ymax = 5.794081279060624; /* image dimensions */ /* draw figures */ draw((-2.5073210715321177,4.1011679907946705)--(-4.8,-1.42), linewidth(2)); draw((-2.5073210715321177,4.1011679907946705)--(7.8,-1.44), linewidth(2)); draw((-4.8,-1.42)--(7.8,-1.44), linewidth(2)); draw(circle((1.5010001557888868,-0.7999018530016957), 6.33143938414973), linewidth(2)); draw((-2.5073210715321177,4.1011679907946705)--(-2.5160905846175696,-1.4236252530402895), linewidth(2)); draw((-2.5073210715321177,4.1011679907946705)--(5.509321383109892,-5.700971696798062), linewidth(2)); draw((5.509321383109892,-5.700971696798062)--(5.516090584617569,-1.4363747469597103), linewidth(2)); draw((5.509321383109892,-5.700971696798062)--(7.8,-1.44), linewidth(2)); /* dots and labels */ dot((-2.5073210715321177,4.1011679907946705),dotstyle); label("$A$", (-2.427604428369967,4.304191085959103), NE * labelscalefactor); dot((-4.8,-1.42),dotstyle); label("$B$", (-4.72451847606814,-1.2208183801256984), NE * labelscalefactor); dot((7.8,-1.44),dotstyle); label("$C$", (7.877469407248865,-1.2415112994743305), NE * labelscalefactor); dot((-2.5160905846175696,-1.4236252530402895),linewidth(4pt) + dotstyle); label("$D$", (-2.427604428369967,-1.2622042188229627), NE * labelscalefactor); dot((5.516090584617569,-1.4363747469597103),linewidth(4pt) + dotstyle); label("$E$", (5.601248278899324,-1.2622042188229627), NE * labelscalefactor); dot((1.5010001557888868,-0.7999018530016957),linewidth(4pt) + dotstyle); label("$O$", (1.5868219252646782,-0.6414166383639963), NE * labelscalefactor); dot((5.509321383109892,-5.700971696798062),linewidth(4pt) + dotstyle); label("$P$", (5.601248278899324,-5.545638523989831), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Since $\angle{ABC}=\beta$ and $\angle{ACB}=\theta$, by Trigonometric Relations in the triangles $\triangle{ABD}, \triangle{ADC}$ and $\triangle{APC}$, respectively (and easy angle markings): $tg\beta=\dfrac{AD}{BD}\implies BD=\dfrac{cos\beta\cdot{AD}}{sin\beta}\implies EC=\dfrac{cos\beta\cdot{AD}} {sin\beta}$ $sin\theta=\dfrac{AD}{AC}\implies AD=sin\theta\cdot{AC}\implies EC=\dfrac{cos\beta\cdot{sin\theta}\cdot{AC}}{sin \beta}$ $tg\beta=\dfrac{AC}{PC}\implies AC=\dfrac{sin\beta\cdot{PC}}{cos\beta}\implies \boxed{EC=sin\theta\cdot{PC}} $ Now define $E'$ such that $E'P\perp{BC}$. In the same way, let's calculate $E'C$, then, by Metric Relations in $\triangle{PE'C}$: $sin\theta=\dfrac{E'C}{PC}\implies E'C=sin\theta\cdot{PC}=EC \implies \boxed{E'\equiv{E}}$ Therefore, $PE\perp{BC}$, $q.e.d!$

random hso geo The key claim is that $\triangle ECP \sim \triangle DAC$ (i think this is the right order), which finishes. Indeed, $\angle BCA' = \angle BAA' = 90 - \angle C = \angle DAC$, and if $H$ is the orthocenter then $CE/CA' = BD/CA' = BD/BH = CA/CD$ by orthocenter reflection, as desired.

Let $H, M$ be the orthocenter of $\Delta ABC$and midpoint of $BC$.It's well known that $P$ is the reflection of $H$ over $M$, and $BD=CE\implies M$ is the midpoint of $DE$, thus $HEPD$ is a paralelogram and we are done. $_{\blacksquare}$

Define $E'$ as the foot of the perpendicular from $P$ to $BC$. And call $M$ the midpoint of $BC$. We want to show that $E'=E$ and that's $\iff \overline{CE'}=\overline{BD} \iff \overline{MD}=\overline{ME'}$, so we just want to show that $(D,E';M,{P_{\infty}}_{BC})=-1$, but we have that $(A,P;O,{P_{\infty}}_{AP})=-1$ (since $\overline{OA}=\overline{OP}$), so projecting through ${P_{\infty}}_{AD}={P_{\infty}}_{OM}={P_{\infty}}_{PE'}$ on $BC$ we have $A \rightarrow D$, $O \rightarrow M$, $P \rightarrow E'$ and ${P_{\infty}}_{AP} \rightarrow {P_{\infty}}_{BC}$ so we get $(D,E';M,{P_{\infty}}_{BC})=-1\ \blacksquare$

hukilau17 wrote: $$\frac{e-p}{b-c} = \frac{(a+b)(a+c)}{2a(b-c)} \in i\mathbb{R}$$$\blacksquare$ how do i know whether the final expression is real or not?

lksb wrote: hukilau17 wrote: $$\frac{e-p}{b-c} = \frac{(a+b)(a+c)}{2a(b-c)} \in i\mathbb{R}$$$\blacksquare$ how do i know whether the final expression is real or not? A complex number is real iff it is its own conjugate.

Actually trivial?? Set up coordinates such that $BC$ is vertical and $O$, the centre of $\Gamma$, is the origin. Then the fact that $P = -A$ means they have opposite y-coordinates, which finishes by projecting onto $BC$.

WLOG suppose that $BC$ isn't a diameter of $O$. Move $A$ with degree $2$ along $(ABC)$. Then $D$ and thus $E$ move with degree $2$, similarly $P$ moves with degree $2$. We want to show that $PE'$ passes through $\infty_{\perp BC}$ so it remains to check $5$ cases. If $BCDE$ is an inscribed rectangle in $(ABC)$, then $A = B, C, D, E$ can be checked to work. Taking $A$ as the arc midpoint of $BC$ gives $2$ more cases so we are done.

Draw the $A$-isosceles trapezoid point.

reflect A and P over the perpendicular bisector of BC to A' and P' respectively, and since AP is a diameter, AA'PP' is a rectangle and AP' and BC are perpendicular so D is the intersection of BC and AP', and the same holds true for E and A'P this is literally trivial 1 minute solve bruh