parmenides51 wrote: Altitudes $AD$ and $BE$ of an acute triangle $ABC$ intersect at $H$. Let $P$ be the point of tangency of the circle with radius $HE$ centered at $H$ with its tangent line going through point $C$, and $Q$ be the point of tangency of the circle with radius $BE$ centered at $B$ with its tangent line going through point $C$. Prove that points $D, P$, and $Q$ are collinear. Is there a typo somewhere ? Geogebra says otherwise, and also if $C$ is inside the circle centered at $H$ with radius $HE$, there couldn't be any tangent. And also, if $C$ is outside the circle, there are 2 tangents, which $P$ does we use ?

Something is lost during the official English translation. What is missing is that both circles are tangent to $AC$ at point $E$ ($P \ne E, Q \ne E$ ). edit: wording has been updated

Let $CF$ be the perpendicular from C to $AB$, and $Q'$ be the intersection of $FD$ with the circumcircle of $BFC$. We have $BFHD$, and $BFECQ'$ are cyclic, thus we have $$ \measuredangle FBH =\measuredangle FDH = \measuredangle FBE = \measuredangle FQ'E \Leftrightarrow HD \parallel EQ' $$Since $HD \parallel EQ$, we must have $Q'=Q$. Let $P'$ be the intersection of $FD$ with the circumcircle of $CHE$, we have $$ \measuredangle HEP' = \measuredangle HDF = \measuredangle EDH = \measuredangle EP'H \Leftrightarrow HE = HP'$$Since $HE=HP=HP'$ and $ \measuredangle HP'C = 90^{\circ}$, thus $C_1(H,HE) $ is tangent to $P'C$ at $P'$, thus $P'=P$. Hence $F,P,D,Q$ are collinear.

Let $P',Q'$ be the midpoints of $EQ,EP$. Obviously, $P',Q'$ are foots of altitude from $E$ to $CH,CD$ respectively. So, $P'Q'$ is Simsom line of $E$ WRT. $CDH$ which always pass through midpoint of $ED$ (since $D$ is orthocenter of $CDH$). Hence, midpoint of $EQ,EP,ED$ are collinear. $D,P,Q$ are colinear.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -25.8897788524355, xmax = 26.825425819370253, ymin = -16.294712095379186, ymax = 15.2243006196171; /* image dimensions */ /* draw figures */ draw(circle((-3.5924722663398616,4.472526816923287), 2.442332155337406), linewidth(0.4) + red); draw(circle((-6.22,0.06), 7.577923153579024), linewidth(0.4) + red); draw((-3.562501958412277,7.297228339098186)--(-6.22,0.06), linewidth(0.4)); draw((-6.22,0.06)--(8.86,-0.1), linewidth(0.4)); draw((8.86,-0.1)--(-3.562501958412277,7.297228339098186), linewidth(0.4)); draw(circle((8.86,-0.1), 13.038676339201421), linewidth(0.4) + linetype("4 4") + red); draw((-3.997833626410152,2.06406911026597)--(8.86,-0.1), linewidth(0.4)); draw((8.86,-0.1)--(-2.4819208844960157,-6.531789123430908), linewidth(0.4)); draw((-3.562501958412277,7.297228339098186)--(-3.6395800146508037,0.032621538616984656), linewidth(0.4)); draw((-3.997833626410152,2.06406911026597)--(-2.4819208844960157,-6.531789123430908), linewidth(0.4)); draw((-3.5924722663398616,4.472526816923287)--(-2.3428993398439237,6.570991460028678), linewidth(0.4)); draw((-3.5924722663398616,4.472526816923287)--(-3.997833626410152,2.06406911026597), linewidth(0.4)); draw((-6.22,0.06)--(-3.5924722663398616,4.472526816923287), linewidth(0.4)); draw((-6.22,0.06)--(-2.4819208844960157,-6.531789123430908), linewidth(0.4)); draw((-4.739492486329917,0.04429169746769143)--(-3.997833626410152,2.06406911026597), linewidth(0.4)); draw((-4.739492486329917,0.04429169746769143)--(-2.4819208844960157,-6.531789123430908), linewidth(0.4)); draw((-3.997833626410152,2.06406911026597)--(-2.3428993398439237,6.570991460028678), linewidth(0.4)); draw((-3.5924722663398616,4.472526816923287)--(8.86,-0.1), linewidth(0.4)); /* dots and labels */ dot((-3.562501958412277,7.297228339098186),dotstyle); label("$A$", (-3.420439002116208,7.654232063613625), NE * labelscalefactor); dot((-6.22,0.06),dotstyle); label("$B$", (-6.069962996717411,0.39384813035574656), NE * labelscalefactor); dot((8.86,-0.1),dotstyle); label("$C$", (9.001355310234887,0.25621052024659247), NE * labelscalefactor); dot((-3.5924722663398616,4.472526816923287),linewidth(4pt) + dotstyle); label("$H$", (-3.454848404643496,4.763842251321389), NE * labelscalefactor); dot((-3.6395800146508037,0.032621538616984656),linewidth(4pt) + dotstyle); label("$D$", (-3.4892578071707847,0.3250293253011695), NE * labelscalefactor); dot((-2.3428993398439237,6.570991460028678),linewidth(4pt) + dotstyle); label("$E$", (-2.2161099136611155,6.862815805485989), NE * labelscalefactor); dot((-3.997833626410152,2.06406911026597),linewidth(4pt) + dotstyle); label("$P$", (-3.8677612349709567,2.3551840744111923), NE * labelscalefactor); dot((-2.4819208844960157,-6.531789123430908),linewidth(4pt) + dotstyle); label("$Q$", (-2.353747523770269,-6.247166557410939), NE * labelscalefactor); dot((-4.739492486329917,0.04429169746769143),linewidth(4pt) + dotstyle); label("$D'$", (-4.590358688044012,0.3250293253011695), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Haven't noticed a solution using inversion, so this post should fix that First notice that $CDPHE$ is a cyclic pentagon. This we see easily since $\angle CDH = \angle CPH = \angle CEH = 90$. Let $\varphi$ be an inversion around $(C,CE)$, from this define $D' = \varphi (D)$. Since we have that $CDPHE$ is a cyclic pentagon, then we must have that $D',P,E$ are colinear. By quick angle chasinge we have that $\angle PCE = 180-2\alpha$, which in turn would give us $\angle D'EC = \angle PEC = \alpha$, giving us $D'E \parallel AB$. This gives us $\angle CD'P = \beta$ By a quick angle chase we have that $\angle PCQ = 2\gamma - 180 +2\alpha = 180-2\beta$ and since $CPQ$ is an isoceles triangle we have that $\angle CQP = \beta$. Thus we have that $CQD'P$ is cyclic, and under our inversion $\varphi$ we have that $P,D,Q$ are colinear points.