equal or supplementary angles in a trapezoid (by Sharygin)

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equal or supplementary angles in a trapezoid (by Sharygin)

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#1 h Aug 5, 2018, 10:58 AM • 3 Y

Y by centslordm, Adventure10, Mango247

In trapezoid $ABCD$ , diagonal $AC$ is equal to leg $BC$ and $H$ is the midpoint of base $AB$ . A variable line through $H$ intersects line $AD$ at $P$ and line $BD$ at $Q$ . Show that angles $ACP$ and $QCB$ are either equal or supplementary.

(I. Sharygin)

This post has been edited 1 time. Last edited by parmenides51, Aug 5, 2018, 11:02 AM
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#2 h Jul 6, 2021, 12:07 AM • 1 Y

Y by centslordm

$[asy]import graph; size(20cm); 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 = -44.424648340905016, xmax = 54.937042248517905, ymin = -19.19643441484827, ymax = 40.348838645739065; /* image dimensions */ pen qqttcc = rgb(0,0.2,0.8); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffqqff = rgb(1,0,1); draw(arc((14.237379324120631,-3.6140714196298322),2.6902984816630755,112.10305544823095,127.70035902047832)--(14.237379324120631,-3.6140714196298322)--cycle, linewidth(2) + qqwuqq); draw(arc((14.237379324120631,-3.6140714196298322),2.6902984816630755,55.059795709476305,70.65709928172367)--(14.237379324120631,-3.6140714196298322)--cycle, linewidth(2) + qqwuqq); /* draw figures */ draw((19.102204174180255,10.244413996835206)--(14.237379324120631,-3.6140714196298322), linewidth(2) + qqttcc); draw((8.710840772757436,9.994069911956498)--(14.237379324120631,-3.6140714196298322), linewidth(2)); draw((8.710840772757436,9.994069911956498)--(19.102204174180255,10.244413996835206), linewidth(2) + qqttcc); draw((33.43969874521158,-66.97812386564989)--(38.14638317217343,-81.89742544539777), linewidth(2)); draw(circle((38.14638317217343,-81.89742544539777), 15.644118317197451), linewidth(2)); draw((26.20303735462521,-71.79322193374351)--(38.14638317217343,-81.89742544539777), linewidth(2)); draw((26.20303735462521,-71.79322193374351)--(33.43969874521158,-66.97812386564989), linewidth(2)); draw((-10.317852701716532,-4.205645099908482)--(8.710840772757436,9.994069911956498), linewidth(2) + qqttcc); draw((19.102204174180255,10.244413996835206)--(26.436412768416016,13.846709806853816), linewidth(2)); draw((5.5453783661019145,7.631918032450054)--(14.237379324120631,-3.6140714196298322), linewidth(2)); draw((14.237379324120631,-3.6140714196298322)--(26.436412768416016,13.846709806853816), linewidth(2)); draw((-10.317852701716532,-4.205645099908482)--(14.237379324120631,-3.6140714196298322), linewidth(2) + qqttcc); draw((-10.317852701716532,-4.205645099908482)--(19.102204174180255,10.244413996835206), linewidth(2)); draw((14.237379324120631,-3.6140714196298322)--(13.906522473468845,10.119241954395852), linewidth(2) + ffqqff); draw((-36.35521747974349,-4.832925660331936)--(-10.317852701716532,-4.205645099908482), linewidth(2)); draw((8.710840772757436,9.994069911956498)--(22.67129707116638,20.411732119765148), linewidth(2)); draw((22.67129707116638,20.411732119765148)--(19.102204174180255,10.244413996835206), linewidth(2)); draw((-36.35521747974349,-4.832925660331936)--(26.436412768416016,13.846709806853816), linewidth(2)); /* dots and labels */ dot((19.102204174180255,10.244413996835206),dotstyle); label("$B$", (19.33542567450987,6.989137473241342), NE * labelscalefactor); dot((14.237379324120631,-3.6140714196298322),dotstyle); label("$C$", (14.582565023571773,-7.359121095574884), NE * labelscalefactor); dot((8.710840772757436,9.994069911956498),dotstyle); label("$A$", (6.780699426748853,10.307172267280093), NE * labelscalefactor); dot((-10.317852701716532,-4.205645099908482),dotstyle); label("$D$", (-9.809474543506779,-6.731384783189174), NE * labelscalefactor); dot((33.43969874521158,-66.97812386564989),dotstyle); label("$B$", (-44.424648340905016,40.348838645739065), NE * labelscalefactor); dot((38.14638317217343,-81.89742544539777),dotstyle); label("$C$", (-44.424648340905016,40.348838645739065), NE * labelscalefactor); dot((26.20303735462521,-71.79322193374351),dotstyle); label("$A$", (-44.424648340905016,40.348838645739065), NE * labelscalefactor); dot((25.121934286122357,-90.56357621359435),linewidth(4pt) + dotstyle); dot((13.906522473468845,10.119241954395852),linewidth(4pt) + dotstyle); label("$H$", (11.802589925853262,10.217495651224992), NE * labelscalefactor); dot((5.5453783661019145,7.631918032450054),dotstyle); label("$P$", (3,7.8)); dot((26.436412768416016,13.846709806853816),linewidth(4pt) + dotstyle); label("$Q$", (27.765027583720844,13.087147364988237), NE * labelscalefactor); dot((1.9597633112020496,-3.909858259769157),linewidth(4pt) + dotstyle); label("$M$", (2.1175153918661893,-6.910738015299377), NE * labelscalefactor); dot((22.67129707116638,20.411732119765148),linewidth(4pt) + dotstyle); label("$X$", (19.60445552267618,19.812893569120842), NE * labelscalefactor); dot((-36.35521747974349,-4.832925660331936),linewidth(4pt) + dotstyle); label("$L$", (-35.99504643169405,-4.130762917591234), NE * labelscalefactor); dot((9.225539128564181,8.72671460377256),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
From $H$ is midpoint of $AB$ and from $AC = BC$ , we have $\angle CHB = 90^{\circ}$ . Now, let's let $M$ be the midpoint of $DC$ , $X = AD \cap BC$ , $L = PQ \cap CD$ , and $T = AB \cap CQ$ . Since $AC, DB,$ and $MH$ are concurrent, we have $(L, M; D,C) \stackrel{X}{=} (L,H;P,Q) \stackrel{C}{=} (A,B;H,T) = -1.$ Since $AT \perp CH$ , $CH$ is angle bisector of $\angle PCQ$ . But from earlier, we had $\angle ACH = \angle BCH$ , so $\angle ACP = \angle QCB$ if $AP < DP$ . Similarly, the two angles are supplementary if $AP > DP.$ $\quad$ $\blacksquare$

This post has been edited 2 times. Last edited by JustinLee2017, Jul 6, 2021, 4:13 PM

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