Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
No topics here!
4 lines concurrent
Zavyk09   7
N May 2, 2025 by bin_sherlo
Source: Homework
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
7 replies
Zavyk09
Apr 9, 2025
bin_sherlo
May 2, 2025
4 lines concurrent
G H J
Source: Homework
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Zavyk09
13 posts
#1 • 1 Y
Y by PikaPika999
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aidenkim119
33 posts
#2 • 1 Y
Y by PikaPika999
............
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aidenkim119
33 posts
#3 • 1 Y
Y by PikaPika999
First three are trivial by pascal, but AD looks a bit hard / '
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ItzsleepyXD
130 posts
#4 • 1 Y
Y by PikaPika999
Redefine
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $A'$ is antipode of $A$ . $O'$ is circumcenter of $(BOC)$ . Point $E,F$ satisfied $OE // A'F // AB , OF // A'E // AC$ then prove $OH,O'A',BE,CF$ concurrent .

Let $B',C'$ be antipode of $B,C$ respectively.
MMP: Fix $(O),B,C$ . Move $A$ on $(O)$ deg 2.
Since $A',E,C'$ collinear and $A',F,B'$ collinear
By $\angle C'EO = \angle BAC = \angle BC'C = \angle C'BO$ so $C',B,O,E$ concyclic.
implies that $E$ deg 2. Also $F$ deg 2.
So line $BE,CF$ deg 1.
$H=$ reflection of $A \infty_{\perp BC} \cap (O)$ across $BC$
Since $H,A'$ deg 2. implies that line $O'A'.OH$ deg 2.
We want to prove $BE,CF,OH$ concurrent first and $BE,CF,O'A'$ concurrent.
but both have deg 1+1+2+1 = 5 .

choose $A= B,C,B',C'$ and midpoint of arc $BC$
the rest of problem is easy. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pingupignu
49 posts
#5 • 1 Y
Y by PikaPika999
My solution may not be elegant but here's some DDIT spam for you guys to enjoy :blush:.
Let $X = KE \cap LF$. I will show that $X, A, D$ and $X, O, H$ are collinear.

Part 1:
I first claim that $\angle \infty_{CH}XL = \angle \infty_{BH}XK$. This follows from
$$\angle \infty_{CH}XL = \angle FLH = \angle FHL = 90^\circ - \angle BHL = 90^\circ - \angle CHK = \angle EHK = \angle XKH = \angle \infty_{BH}XK.$$Then, applying DDIT on $X \cup LDKH$ we see that $$(XK, XL), (XH, XD), (X\infty_{BH}, X\infty_{CH})$$are reciprocal pairs under some involution on $\mathcal{P}_X$. This involution must be a reflection in the angle bisector of $\angle KXL$. Hence $XH, XD$ are isogonal in $\angle KXL$.

Next, since $AK=AL$ (well-known), $LF=FH=AE$, $AF=EH=EK$, we yield $\triangle LAF \cong \triangle AKE$.
I claim that $X\infty_{AB}, X\infty_{AC}$ are isogonal in $\angle XKL$. This is because $$\angle \infty_{AB}XF = 180^\circ - \angle AFL = 180^\circ - \angle AEK = \angle AEX = \angle \infty_{AC}XE.$$Applying DDIT on $X \cup AEHF$ would then give $XA, XH$ are isogonal in $\angle EXF$. Since $XH, XD$, $XH, XA$ are isogonal in $\angle KXL = \angle EXF$ we conclude that $XA \equiv XD$, or $X \in AD$.

Part 2:
I first prove that $X\infty_{CH}, OL, AK$ concur at a point $S$ on $(XLK)$. For this, let $S = X\infty_{CH} \cap OL$, where from a short angle chase we get $$\angle XSL = \angle OLH = B-A = (90^\circ - A) - (90^\circ - B) = \angle AKL - \angle LCB$$$$= \angle AKL - \angle LAF = \angle AKL - \angle  AKE = \angle EKL = \angle XKL$$Hence $XSKL$ are cyclic, and from
$$\angle SKX = \angle SLX = \angle FLO = \angle ALO - \angle ALF = A - \angle EAK = A - (90^\circ - C)$$which equals
$$= A+C-90^\circ = 90^\circ - B = \angle LAB = \angle AKE = \angle AKX$$$\implies S \in AK$. The claim is proven.

From DDIT in $X \cup AKOL$ we have the reciprocal pairs $$(XA, XO), (XK, XL), (XS, XT)$$where $T = AL \cap KO \cap X \infty_{BH}$ (similarly).
since $(XS, XT) = (X\infty_{CH}, X\infty_{BH})$ and we have established $(XA, XH), (X\infty_{CH}, X\infty_{BH})$ are isogonal in $\angle KXL$ we get $XH \equiv XO \implies X \in OH$. The problem is solved. $\blacksquare$
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
tomsuhapbia
6 posts
#6 • 1 Y
Y by Amkan2022
We need two well-known lemmas about the isogonal line:
1. Given a $\triangle ABC$ and a point $P$ satisfied $\angle ABP=\angle ACP$. Let $Q$ be the reflection of $P$ in the midpoint of $BC$. Then $AP$ and $AQ$ are isogonals wrt $\angle BAC$.
2. Given a trapezoid $ABCD$ $(AB\parallel CD)$ is inscribed $(O)$. Let $E,F$ be the intersections of $BC$ and $AD$; $AC$ and $BD$. Let $S$ be a abitary point on $(O)$. Then $SE,SG$ are isogonals wrt $\angle ASB$.

Back to the problem: Let $X,Y$ be the intersections of $OL$ and $AK$; $AL$ and $OK$. By symmetric, we have $AK=AH=AL$ so $AO$ is the perpendicular bisector of $KL$ then $XKLY$ is a trapezoid. Let $LK$ intersects $KE$ at $Z$.

We have
$$\angle FZK=\angle ZEH-\angle ZFC=180^\circ-\angle KEH-\angle HFC=180^\circ-3\angle BAC$$and
$$\angle LYK=\angle AOK-\angle OAL=2(90^\circ-\angle OAL)-\angle OAL=180^\circ-3\angle OAL=180^\circ-3\angle BAC=\angle FZK\,(2)$$since
$$\angle OAL=\dfrac{1}{2}\angle KAL=\dfrac{1}{2}(\angle KAB+\angle BAH+\angle HAC+\angle CAL)=\angle BAC$$$(2)$ leads to $Z$ lies on the circumcircle of $XKLY$. From the lemma 2, we obtain that $ZA,ZO$ are isogonals wrt $\angle LZK$ $(1)$. We also have $\angle ZFE=\angle LFE=\angle KEH=180^\circ-\angle ZEH$ and $\angle HEZ=\angle HKE=\angle FLH=180^\circ-\angle HLZ$, using lemma 1 and get $(XD,XH)$ and $(XH,XA)$ are two isogonal pairs wrt $\angle LZK\equiv\angle FZE$, so $A,D,Z$ are collinear. Combine with $(1)$ and we conclude $Z,O,H$ are collinear or $AD,LF,KE,OH$ are concurrent at $Z$.

https://i.postimg.cc/hJQdTNdW/image.png
This post has been edited 2 times. Last edited by tomsuhapbia, May 1, 2025, 5:25 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hectorleo123
344 posts
#7
Y by
I apologize for the complex bash, but I couldn't find another way.

Let \( B' \) and \( C' \) be the antipodes of \( B \) and \( C \).
Since \( L \) is the reflection of \( H \) over \( AB \), we have \( \angle FLB = \angle FHB = 90^\circ \) (since \( FH \parallel AC \perp BH \)).
Analogously, \( \angle EKC = 90^\circ \).
\(\Rightarrow B', F, L \) are collinear and \( C', E, K \) are collinear.
By Pascal's Theorem on
\[ \binom{B, L, C'}{C, K, B'} \]we get that \( KE, LF \), and \( OH \) are concurrent.
Now it suffices to prove that \( AD, KE \), and \( LF \) are concurrent.

We use complex numbers, where \( (ABC) \) is the unit circle and \( a = 1 \).
Let
\[
k = -\frac{c}{b}, \quad l = -\frac{b}{c}, \quad c' = -c, \quad b' = -b, \quad h = b + c + 1, \quad o = 0
\]\[
d + b + c + 1 = d + h = l + k = -\frac{b}{c} - \frac{c}{b}
\]\[
\Rightarrow d = -\frac{b^2 + c^2 + b^2c + bc^2 + bc}{bc}
\]
Let \( X = KC' \cap LB' \).
We have:
\[
\frac{x + c}{\overline{x + c}} = \frac{c - \frac{c}{b}}{\overline{c - \frac{c}{b}}} = -\frac{c^2}{b}
\Rightarrow \overline{x} = -\frac{xb + bc + c}{c^2}
\]Analogously,
\[
\overline{x} = -\frac{xc + bc + b}{b^2}
\]Equating both expressions:
\[
(b - c)(b^2c + bc^2 + bc - x(b^2 + bc + c^2)) = 0
\Rightarrow x = -\frac{bc(b + c + 1)}{b^2 + bc + c^2}
\]
Points \( A, D, X \) are collinear if and only if
\[
\frac{d - 1}{x - 1} \in \mathbb{R}
\]Substituting:
\[
\frac{(b^2 + c^2 + b^2c + bc^2 + 2bc)/bc}{(b^2c + bc^2 + 2bc + b^2 + c^2)/(b^2 + bc + c^2)} = \frac{b^2 + bc + c^2}{bc}\in \mathbb{R}_\blacksquare
\]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bin_sherlo
719 posts
#8
Y by
Let $B',C'$ be the antipodes of $B,C$ on $(ABC)$. Also let $LC'\cap KB'=W,KC'\cap LB'=P$.
Claim: $B',F,L$ and $C',E,K$ are collinear.
Proof: Pascal at $KB'LCAB$ yields $AC_{\infty},B'L\cap AB,H$ are collinear thus, $B'L\cap AB=F$. Similarily $C',E,K$ are collinear.
Claim: $P$ lies on $OH$.
Proof: Pascal at $BKC'CLB'$ gives $H,P,O$ are collinear.
Claim: $A,D,P$ are collinear.
Proof: Notice that $W,L,H,K$ lie on the circle with diameter $WH$ and since $AH=AK=AL$, $A$ must be the circumcenter of $(KLHW)$. Hence $W,A,H$ are collinear.
DDIT at $DLHK$ implies $(\overline{AD},\overline{AH}),(\overline{AK},\overline{AL}),(\overline{AC'},\overline{AB'})$ is an involution. DDIT at $B'KC'L$ gives $(\overline{AP},\overline{AH}),(\overline{AK},\overline{AL}),(\overline{AC'},\overline{AB'})$ is an involution. Combining these implies $AD\equiv AP$ hence $A,D,P$ are collinear as desired.$\blacksquare$
Z K Y
N Quick Reply
G
H
=
a