Point on circumcircle with angle condition

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Point on circumcircle with angle condition

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Source: OMMC POTM 12/2022

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1757 posts

Taco12

#1 h Dec 3, 2022, 12:06 AM • 4 Y

Y by mathking999, megarnie, Amoonguss, ItsBesi

Let $\triangle ABC$ be such that the midpoint of $BC$ is $D$ . Let $E$ be the point on the opposite side of $AC$ as $B$ on the circumcircle of $\triangle ABC$ such that $\angle DEA = \angle DEC$ and let $\omega$ be the circumcircle of $\triangle CED$ . If $\omega$ intersects $AE$ at $X$ and the tangent to $\omega$ at $D$ intersects $AB$ at $Y$ , show that $XY$ is parallel to $BC$ .

Proposed by Taco12

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#2 h Dec 3, 2022, 12:57 AM

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First, note that by the definition of the circumcircle, E is the midpoint of arc AC on the circle. Since angle DEA = angle DEC, it follows that angle ACD = angle DEC = angle CDE and therefore angle ACD = angle CDE.

Since D is the midpoint of BC, it follows that BD = DC. Let Z be the foot of the perpendicular from C to AB. By the angle bisector theorem, we have BD:DC = BZ:ZC. Since BD = DC, it follows that BZ = ZC. Therefore, Z is the midpoint of AB, so Z lies on the circumcircle of triangle ABC as well.

Since angle ACD = angle CDE, we have triangle ACD ~ triangle CDE. Therefore, AD:DC = CD:DE. Since CD = DE, it follows that AD = DC, so D is the midpoint of AD. Similarly, we have triangle CDE ~ triangle DEC, so DE:EC = CD:DC. Since CD = DC, it follows that DE = EC, so E is the midpoint of DE.

Since E is the midpoint of arc AC on the circle and D is the midpoint of AD, it follows that E is the midpoint of AD as well. Therefore, X is the midpoint of DE, so DX = XE.

Now, consider the tangent line to omega at D. Let this line be denoted by L. Since DX = XE, it follows that D is the midpoint of XE, so L is parallel to XE. Since omega is the circumcircle of triangle CED, it follows that L is perpendicular to DE. Since DE is parallel to AC and L is perpendicular to DE, it follows that L is parallel to BC. Since L is parallel to BC, it follows that Y lies on line BC, so XY is parallel to BC.

Therefore, we have shown that XY is parallel to BC, as desired.

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#3 h Dec 3, 2022, 1:18 AM

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By the angle condition, we have that $XD=DC$ . By Reim's theorem, we have that $XD\parallel AB$ . In particular, we have that $BYXD$ is a parallelogram. Since $BD=DC$ , we have that $XYDC$ is a parallelogram as well, so $\angle XCD = \angle CXD = \angle XDY$ , so we have the desired tangency.

This post has been edited 1 time. Last edited by PROA200, Dec 3, 2022, 1:19 AM

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#4 h Dec 3, 2022, 12:30 PM • 4 Y

Y by megarnie, Mango247, Mango247, Mango247

Copy-pasted my solution from the OMMC Discord.

$\textit{Solution.}$ Note that $\angle XED = \angle AED = \angle CED$ , which means that $D$ is midpoint of the arc $CX$ in $(XECD)$ . Thus, $XD = CD = BD$ . Thus, $\triangle CXB$ is a right-angled triangle. Also, $\angle XDC = 180 - \angle XEC = 180 - \angle AEC = 180 - (180 - \angle B) = \angle B$ , which means that $XD \parallel AB$ . Therefore, we have uniquely defined the point $X$ . Now, let us add the point $Y$ to the picture. Note that $\angle YDX = \angle DCX = 90 - \frac{\angle XDC}{2} = 90 - \frac{\angle B}{2}$ . Now, note that $\angle BDY = 180 - (90 - \frac{\angle B}{2} + \angle B) = 90 - \frac{\angle B}{2}$ . Also, $\angle BYD = 90 - \frac{\angle B}{2}$ . Thus, $BY = BD = DC = DX$ . Thus, $XD \parallel BY$ , and $XD = BY$ , which means that quadrilateral $YXDB$ is a parallelogram. Thus, $XY \parallel BD$ , which means that $XY \parallel BC$ , which completes the proof. $\blacksquare$

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1757 posts

Taco12

#5 h Dec 3, 2022, 11:04 PM • 6 Y

Y by megarnie, mannshah1211, Spectator, samrocksnature, Mango247, Mango247

$[asy] import graph; size(10cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; /* point style */ real xmin = -12.511836068944065, xmax = 23.74797334637722, ymin = -3.3710528306115997, ymax = 17.727860661927153; pen zzttqq = blue; draw((0,0)--(13,0)--(5,14)--cycle, linewidth(0.1) + zzttqq); draw((0,0)--(13,0), linewidth(0.1) + zzttqq); draw((13,0)--(5,14), linewidth(0.1) + zzttqq); draw((5,14)--(0,0), linewidth(0.1) + zzttqq); draw(circle((6.5,5.571428571428571), 8.561005567486252), linewidth(0.1)); draw((xmin, 0.5714285714285714*xmin + 1.8571428571428572)--(xmax, 0.5714285714285714*xmax + 1.8571428571428572), linewidth(0.1)); draw((9,7)--(1.6954206790080075,12.657106652941074), linewidth(0.1)); draw((1.6954206790080075,12.657106652941074)--(5,14), linewidth(0.1)); draw((1.6954206790080075,12.657106652941074)--(0,0), linewidth(0.1)); draw(circle((4.968825343677512,9.33932876781572), 4.660775492737979), linewidth(0.1)); draw((xmin, 1.7232185196809595*xmin-8.508966677128642)--(xmax, 1.7232185196809595*xmax-8.508966677128642), linewidth(0.1)); draw((xmin, -1.7499198853196327*xmin + 8.640813583286176)--(xmax, -1.7499198853196327*xmax + 8.640813583286176), linewidth(0.1)); dot((0,0),dotstyle); label("$A$", (0.12224386670794997,0.3244155505665954), NE * labelscalefactor); dot((13,0),dotstyle); label("$B$", (13.135346200429526,0.3244155505665954), NE * labelscalefactor); dot((5,14),dotstyle); label("$C$", (5.112705441290496,14.316659079301129), NE * labelscalefactor); dot((9,7),linewidth(4pt) + dotstyle); label("$D$", (9.12402582086001,7.241574315336037), NE * labelscalefactor); dot((1.6954206790080075,12.657106652941074),dotstyle); label("$E$", (1.827844658020972,12.958495486218544), NW * labelscalefactor); dot((0.9376506873550242,7),linewidth(4pt) + dotstyle); label("$X$", (1.0697998618818512,7.241574315336037), NE * labelscalefactor); dot((4.937833815007984,0),linewidth(4pt) + dotstyle); label("$Y$", (5.049535041612236,0.26124515088833566), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

From cyclic quadrilateral $XECD$ , we get $\angle DXC = \angle DEC = \angle XED = \angle DCX.$ Call this angle $\theta$ . Now, from the tangency, we get $YD \parallel XC$ . Additionally, we have $\angle YDB = \theta$ , and using cyclic quadrilateral $AECB$ yields $\angle YBD = 180-2\theta$ and $\angle BYD=\theta$ . However, since $\angle YDX = \theta$ , we have $BY \parallel XD$ . By our congruent angles and the given midpoint, we have that $BYXD$ is a parallelogram which implies the needed result. $\blacksquare$

This post has been edited 3 times. Last edited by Taco12, Dec 29, 2024, 7:12 PM

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#6 h Dec 7, 2022, 2:58 PM • 3 Y

Y by BMO2023IMO2023, megarnie, Taco12

Check out v4913's video explanation of this problem on the official OMMC youtube channel:

https://www.youtube.com/watch?v=saTWVb6RT8c

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1934 posts

#7 h Dec 8, 2022, 2:33 AM • 1 Y

Y by teomihai

Connect $XD$ and $XC$ . Also, let $\angle DEA=\alpha$ . Then $\angle DXC=\angle DCX=\alpha$ , so $DX=CD=BD$ . Also, $\angle YDX=\angle DEX=\alpha$ , and $\angle BDY=180^{\circ}-\angle YDX-\angle XDC=\alpha$ . Therefore, $\triangle BDY\cong\triangle XDY$ by SAS, so $\angle BYD=\angle XYD$ . Finally, $\angle ABC=180^{\circ}-\angle AEC=180^{\circ}-2\alpha$ , so $\angle BYD=180^{\circ}-\angle YBD-\angle YDB=\alpha$ and $\angle XYD=\alpha$ because of our congruence from earlier. Then $\angle XYD=\angle BDY$ , so $XY$ and $BC$ are parallel. QED.

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4192 posts

#8 h Apr 6, 2023, 3:51 AM

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Claim 1: $YD\parallel XC$ . Let $\angle AED=\angle CED=\alpha$ . Then, from $XECD$ cyclic, $\angle DXC=\angle DCX=\alpha.$ Furthemore, $\angle YDX=\alpha$ due to the tangency, so $\angle YDB=2\alpha-\alpha=\alpha$ . Since $\angle YDB=\angle XCD,$ $YD\parallel XC.$

Claim 2: $BY\parallel XD.$ Note that $\angle XDC=180-2\alpha$ . However, from cyclic $AECB$ , we also have $\angle YBD=180-\angle AEC=180-2\alpha$ , so $BY\parallel XD.$

Note that then $\triangle YBD\sim \triangle XDC$ . However, $BD=CD$ , so in fact $\triangle YBD\equiv \triangle XDC$ . Hence, the height from $Y$ to $BC$ is the same as the height from $X$ to $BC$ , so we are done.

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#9 h Jul 13, 2023, 4:08 PM • 1 Y

Y by OronSH

Luv so xor

Let $\angle{DEC} = \angle{DEX} = \alpha$ . By tangent properties $\angle{EDY} = \angle{ECD} \implies \angle{BDY} = \alpha$ . Because $ECBA$ is cyclic, $\angle{DBY} = 180-2\alpha$ . Thus, $\angle{DYA} = 180-\alpha$

It suffices to prove that $\angle{DYX} = \alpha$ . Because $\angle{DEC} = \angle{DEX} \implies DB = ED = DX$ . Also note that $\triangle DBY$ is isosceles so $DB = BY$ . This implies that $DX \parallel BY$ because $\triangle CDX \cong \triangle DBY$ . This should be enough to imply that $XY \parallel DB$ .

remarks

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#13 h Apr 1, 2024, 8:43 PM

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Took me long enough

Let $\odot (ABC)=\Gamma , \odot (CED)=\omega$

$\angle ABC + \angle AEC=\stackrel{\Gamma }{=}180 \implies \angle AEC=180-\angle ABC=180-\angle B \implies \angle AEC=180-\angle B$

So $\angle AED=\angle CED=\frac{\angle AEC}{2}=\frac{180-\angle B}{2}=90-\frac{\angle B}{2} \implies \angle AED=\angle CED=90-\frac{\angle B}{2}$

$\angle XCD \stackrel{\omega }{=}\angle XED= \angle AED=\angle CED \stackrel{\omega }{=} \angle CXD=90-\frac{\angle B}{2} \implies \angle XCD=\angle CXD=90-\frac{\angle B}{2} \implies$
The triangle $\triangle CXD$ -is isosceles $\iff DC=DX$

From the triangle $\triangle XCD$ we have:

$\angle XCD+\angle CXD+\angle CXD=180 \implies 90-\frac{\angle B}{2}+90-\frac{\angle B}{2}+\angle CXD=180 \implies \angle CXD=\angle B$

Since $DY$ -is tangent to $\omega$ we get:

$\angle YDX=\angle XCD=90-\frac{\angle B}{2} \implies \angle YDX=90-\frac{\angle B}{2}$

$\angle BDY+ \angle YDX+\angle CDX=\angle BDC=180 \implies \angle BDY+90-\frac{\angle B}{2} +\angle B=180 \implies \angle BDY=90-\frac{\angle B}{2}$

From the triangle $\triangle BDY$ we have:

$\angle BDY+\angle DBY+\angle BYD=180 \implies 90-\frac{\angle B}{2} + \angle B+\angle BYD=180 \implies \angle BYD=90-\frac{\angle B}{2}$

So $\angle BYD=90-\frac{\angle B}{2}=\angle BDY \implies \angle BYD=\angle BDY \implies$ The triangle $\triangle BDY$ -is isosceles $\iff BY=BD \implies BY=BD=DC=DX \implies BY=DX$

Since $\angle BYD=90-\frac{\angle B}{2}=\angle YDC \implies \angle BYD=\angle YDC$ we get that $BY \| DX$

$\textbf{Claim:}$ $XY \| BC$

$\textbf{Proof:}$ Let $AX \cap DY=\{O\}$

Since $BY \| DX$ we get: $\angle BYD=\angle YDX , \angle YBD=\angle BXD$ Combining with $BY=DX$ we get by $ASA$ criterion that the triangles $\triangle BOY , \triangle DOX$ are congruent $\iff \triangle BOY \cong \triangle DOX \implies BO=OX , DO=OY.$ Combining with the fact that $\angle BOD=\angle XOD$ we get again by $SAS$ criterion that triangles $\triangle AOD , \triangle XOY$ are congruent $\iff \triangle BOD \cong \triangle XOY \implies \angle XYO=\angle BDO \implies \angle XYD \equiv \angle XYO =\angle BDO \equiv \angle BDY \implies XY \| BD \implies XY \| BC$ . As desired $\blacksquare$

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#14 h Jun 7, 2025, 4:44 AM

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Parallel lines ftw!!!

We will use undirected angles because I we are lazy.

Claim: $\triangle{DCX}$ is isosceles.

Proof of Claim: Evident from the angle bisector condition.

In particular, this implies $BD=CD=XD$ .

Claim: $DY||CX$ .

Proof of Claim: This is pretty clear with angle chasing on $DCX$ that I am too lazy to do because it is not particularly interesting.

Claim: $DX||BY$ .

Proof of Claim: $\angle{CEX}=180-\angle{B}$ , and $\angle{XDC}=180-\angle{CEX}$ implies $XDC=\angle{B}$ , which implies the result by CAP.

This also implies that $\angle{BDY}=\angle{XDY}$ by further angle chasing.

In particular, $DY$ is perpendicular to $BX$ , which means that $BDXY$ is a kite. But $DX||BY$ implies that it is a rhombus, and $XY||BD$ implies $XY||BC$ which is what we wanted.

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