DE = QF if CD = DF, PF //AC ,PQ // CD, ABC isosceles

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DE = QF if CD = DF, PF //AC ,PQ // CD, ABC isosceles

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Source: 2020 Kharkiv Lyceum No 27 10.4

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#1 h Nov 2, 2021, 6:24 PM

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On the lateral side $AB$ of the isosceles triangle $ABC$ , we mark the point $D$ . On the lateral side $AC$ , we mark the point $E$ and on the extension of the base $BC$ beyond $B$ , we mark the point $F$ such that $CD = DF$ . On the line $DE$ , the point $P$ is chosen, and on the segment $BD$ , the point $Q$ is chosen so that $PF \parallel AC$ and $PQ \parallel CD$ . Prove that $DE = QF$ .

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#2 h Nov 2, 2021, 9:25 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -18.522898386334575, xmax = 16.308307897325022, ymin = -5.982661310412525, ymax = 14.84330798190614; /* image dimensions */ /* draw figures */ draw((0,7.02)--(-3.06,0), linewidth(0.4) + blue); draw((-3.06,0)--(3.06,0), linewidth(0.4) + blue); draw((0,7.02)--(3.06,0), linewidth(0.4) + blue); draw((-1.5628729281767955,3.4345856353591158)--(-6.18574585635359,0), linewidth(0.4) + blue); draw((-6.18574585635359,0)--(-3.06,0), linewidth(0.4) + blue); draw((-1.5628729281767955,3.4345856353591158)--(3.06,0), linewidth(0.4) + blue); draw((-1.5628729281767955,3.4345856353591158)--(2.1581546961325966,2.068939226519337), linewidth(0.4) + blue); draw((-8.848394562792178,6.1084293853591145)--(-6.18574585635359,0), linewidth(0.4) + blue); draw((-8.848394562792178,6.1084293853591145)--(-1.5628729281767955,3.4345856353591158), linewidth(0.4) + blue); draw((-6.18574585635359,0)--(-2.464718232044199,1.3656464088397782), linewidth(0.4) + blue); draw(circle((-5.515722546389463,3.926597007400534), 3.983352268431968), linewidth(0.4) + linetype("4 4") + red); draw((-8.848394562792178,6.1084293853591145)--(-2.464718232044199,1.3656464088397782), linewidth(0.4) + blue); /* dots and labels */ dot((0,7.02),dotstyle); label("$A$", (0.09770079533462242,7.249559353484268), NE * labelscalefactor); dot((-3.06,0),dotstyle); label("$B$", (-2.971628740105355,0.22420508347720797), NE * labelscalefactor); dot((3.06,0),linewidth(4pt) + dotstyle); label("$C$", (3.1442945564380076,0.17873353480402313), NE * labelscalefactor); dot((-1.5628729281767955,3.4345856353591158),dotstyle); label("$D$", (-1.471067633890255,3.6573070083026646), NE * labelscalefactor); dot((-6.18574585635359,0),linewidth(4pt) + dotstyle); label("$F$", (-6.086429824218518,0.17873353480402313), NE * labelscalefactor); dot((2.1581546961325966,2.068939226519337),dotstyle); label("$E$", (2.257599357310903,2.293160548107119), NE * labelscalefactor); dot((-8.848394562792178,6.1084293853591145),linewidth(4pt) + dotstyle); label("$P$", (-8.746515421599833,6.294656831347386), NE * labelscalefactor); dot((-2.464718232044199,1.3656464088397782),linewidth(4pt) + dotstyle); label("$Q$", (-2.3804986073539522,1.5428799949995688), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $\alpha = \ angle ABC = \angle BCA$ , and let $x= \angle DCF = \angle DFC$ .

Since we have that $PF \parallel AC$ and $PQ \parallel DC$ , by angle chasing we get that $\angle QPF = \angle DCE = \alpha - x$ .
Notice that $\angle FDQ = 180-2x-180+\alpha+x=\alpha-x$ , this implies that $FPDQ$ is cyclic, from which we get that $FDQ \cong DCE$ , which gives us that $FQ=DE$ .

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