collinear wanted, altitudes , orthocenter, 2 circles with diameters

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Source: 2016 NZOMC Camp Selections p6

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#1 h Jan 10, 2021, 9:49 AM

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Altitudes $AD$ and $BE$ of an acute triangle $ABC$ intersect at $H$ . Let $P \ne E$ be the point of tangency of the circle with radius $HE$ centred at $H$ with its tangent line going through point $C$ , and let $Q \ne E$ be the point of tangency of the circle with radius $BE$ centred at $B$ with its tangent line going through $C$ . Prove that the points $D, P$ and $Q$ are collinear.

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#2 h Jan 10, 2021, 12:18 PM

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$[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 = -33.4033052656582, xmax = 23.61085238263149, ymin = -21.872386618234476, ymax = 13.752145873115111; /* image dimensions */ /* draw figures */ draw(circle((-3.2554735257759493,-1.0281510994642318), 7.1138897416703815), linewidth(0.8) + red); draw((-0.07868818964700604,5.337024639256514)--(-10.14,-2.82), linewidth(0.8) + blue); draw((-10.14,-2.82)--(3.18,-4.06), linewidth(0.8) + blue); draw((3.18,-4.06)--(-0.07868818964700604,5.337024639256514), linewidth(0.8) + blue); draw((-0.07868818964700604,5.337024639256514)--(-0.9179719268020698,-3.6785071179253324), linewidth(0.8) + blue); draw((-10.14,-2.82)--(1.3662988726427505,1.170139611445795), linewidth(0.8) + blue); draw(circle((-0.5277411380951058,0.5133268381849743), 2.00469214130111), linewidth(0.8) + red); draw(circle((-10.14,-2.82), 12.178510905090517), linewidth(0.8) + red); draw((-1.5620298200838647,-1.2039493491839606)--(3.18,-4.06), linewidth(0.8) + blue); draw((3.18,-4.06)--(0.432051855645617,-8.865481571136305), linewidth(0.8) + blue); draw((-1.5620298200838647,-1.2039493491839606)--(0.432051855645617,-8.865481571136305), linewidth(0.8) + blue); draw(circle((3.18,-4.06), 5.535690754963768), linewidth(0.8) + linetype("4 4") + red); draw((-1.5620298200838647,-1.2039493491839606)--(1.3662988726427505,1.170139611445795), linewidth(0.8) + blue); draw((1.3662988726427505,1.170139611445795)--(0.432051855645617,-8.865481571136305), linewidth(0.8) + blue); draw((-0.5277411380951058,0.5133268381849743)--(-1.5620298200838647,-1.2039493491839606), linewidth(0.8) + blue); draw((-10.14,-2.82)--(0.432051855645617,-8.865481571136305), linewidth(0.8) + blue); draw(circle((1.3261294309524472,-1.7733365809075132), 2.943750274564138), linewidth(0.8) + linetype("4 4") + red); draw((-0.5277411380951058,0.5133268381849743)--(3.18,-4.06), linewidth(0.8) + blue); /* dots and labels */ dot((-0.07868818964700604,5.337024639256514),dotstyle); label("$A$", (0.05990621680202132,5.703088322768343), NE * labelscalefactor); dot((-10.14,-2.82),dotstyle); label("$B$", (-10.001415721131453,-2.4577616935554643), NE * labelscalefactor); dot((3.18,-4.06),dotstyle); label("$C$", (3.33915188546182,-3.6874788193028873), NE * labelscalefactor); dot((-0.5277411380951058,0.5133268381849743),linewidth(4pt) + dotstyle); label("$H$", (-0.3872636471061331,0.8214839751043304), NE * labelscalefactor); dot((-0.9179719268020698,-3.6785071179253324),linewidth(4pt) + dotstyle); label("$D$", (-0.7599052003629284,-3.3893655766974513), NE * labelscalefactor); dot((1.3662988726427505,1.170139611445795),linewidth(4pt) + dotstyle); label("$E$", (1.5132082745035231,1.4549746156408816), NE * labelscalefactor); dot((-1.5620298200838647,-1.2039493491839606),linewidth(4pt) + dotstyle); label("$P$", (-1.43065999622516,-0.8926671698769258), NE * labelscalefactor); dot((0.432051855645617,-8.865481571136305),linewidth(4pt) + dotstyle); label("$Q$", (0.5816043913615347,-8.5690831669669), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Nice exercise

$\color{black}\rule{25cm}{1pt}$
By the power of a point we must have that $CP^2=CE^2=CQ^2$ . This implies that $C$ is the center of $(PEQ)$ .
This implies that $\angle CPQ = \angle CQP$ .

Since we have that $\angle HPC = \angle HDC = 90$ , we have that $CDPHE$ is a cyclic quadrilateral. Thus we have that $\angle DPC = \angle DHC = \angle B$ .
This implies that we must show that $\angle PCQ = 180 -2\angle B$ .

Since we have that $\angle BCE = \angle BCQ = \angle C$ and since we have that $\angle PCD = \angle C - 180 + \angle PHE$ , we want that $\angle PHE = 2 \angle A$ .
But this is easily seen since we have that $\angle PHE = 2\angle EHC = 2 \angle A$ .
Thus we have that $\angle CPQ \equiv \angle B$ , this implies that $P,D$ and $Q$ are colinear points.

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#3 h Jan 10, 2021, 12:22 PM • 1 Y

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here
I answered there.

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