Line through orthocenter

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Line through orthocenter

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Source: Mexico National Olympiad 2011 Problem 2

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323 posts

#1 h Jun 22, 2014, 4:09 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute triangle and $\Gamma$ its circumcircle. Let $l$ be the line tangent to $\Gamma$ at $A$ . Let $D$ and $E$ be the intersections of the circumference with center $B$ and radius $AB$ with lines $l$ and $AC$ , respectively. Prove the orthocenter of $ABC$ lies on line $DE$ .

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3212 posts

#2 h Jun 23, 2014, 2:05 AM • 2 Y

Y by Adventure10, Mango247

Let $BL$ be the altitude of triangle $ABC$ with circumcenter $O$ . Hence $BC^2-AB^2=LC^2-LA^2$ .
By Power of a Point, we have $CE \cdot CA=CB^2-AB^2=LC^2-LA^2$ . It follows that $LE=LA$ or $L$ is midpoint of $AE$ .
$BL \cap DE=H'$ . Since $L$ is the midpoint of $AE$ then $H'AE$ is isosceles triangle with $H'A=H'E$ . We also have $\angle H'EA= \frac{ \angle ABD}{2}= 90^{\circ}- \angle DAB= \angle BAO.$
It follows that $\triangle H'AE \sim \triangle AOB \; ( \text{A.A})$ . Let $X$ be the midpoint of $AB$ then $OX \perp AB$ . Hence $\frac{AE}{AB}= \frac{H'L}{OX}$ or $\frac{AL}{AB}= \frac{H'L}{2OX}$ .
Let $H$ be the othorcenter of triangle $ABC$ . We have $H \in BL$ and $\frac{AL}{AB}= \frac{HL}{HC}= \frac{HL}{2OX}$ . Thus, $LH=LH'$ . We also have $H,H'$ are between $B$ and $L$ so $H \equiv H'$ . That means $H \in DE$ . $\blacksquare$

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111 posts

#3 h Jun 23, 2014, 4:01 AM • 2 Y

Y by Adventure10, Mango247

please give a picture for problem

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216 posts

#4 h Jun 23, 2014, 7:13 AM • 2 Y

Y by Adventure10, Mango247

Another way:

Lemma : Line d passes through the orthocenter of triangle ABC if and only if the

intersection point of the reflections of d about segments AB and AC lies on the

circumcircle of triangle ABC.

Let the other intersection point of circles Γ and W(B,AB) be P.

→ AB = BE ⇒ ∠ABE = 180 - 2∠BAC ⇒ ∠ADE = 90 - ∠BAC ⇒ ∠AED = 90 - ∠ACB

⇒ ∠AED = 90 - ∠ACB (*)

→ ∠EBC = ∠ABC - 180 + 2∠BAC = ∠BAC - ∠ACB

→ ∠ABO = ∠OBP ⇒ 90 - ∠ACB = 90 - ∠BAC + ∠CBP ⇒ ∠CBP = ∠BAC - ∠ACB

⇒ ∠CBP = ∠EBC ⇒ BC ⊥ PE ⇒ ∠CEP = 90 - ∠ACB = ∠AED

⇒ PE is the reflection of DE about AC.

→ Let L be the intersection point of DE and AB : ∠ALE = 180 - (∠BAC + 90 - ∠ACB)

⇒ ∠ALE = 90 - ∠BAC + ∠ACB = ∠BPE ⇒ ∠BLP = ∠BEP = ∠BPE = ∠ALE

⇒ PL is the reflection of DE about AB.

⇒ By the lemma, DE passes through the orthocenter of triangle ABC.

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9795 posts

jayme

#5 h Jun 23, 2014, 12:17 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
you can se my proof on
http://perso.orange.fr/jl.ayme vol. 18 Regard 1, p. 6-8
Sincerely
Jean-Louis

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87 posts

#6 h Jun 23, 2014, 1:07 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Let $H$ be the orthocenter of $\bigtriangleup ABC$ . Then $BH$ is the perpendicular bisector of $AE$ , thus $\angle AEH=\angle HAC=90^\circ-\angle C=90^\circ-\angle DAB=\frac{\angle DBA}{2}=\angle AED$ Since $\angle C>90^\circ$ , clearly $D, B$ and $B,H$ lie on the same side of $AC$ . Thus so does $D,H$ , and so $D,H,E$ are collinear.

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143 posts

#7 h Sep 27, 2019, 7:29 AM • 2 Y

Y by Adventure10, Mango247

Let $H$ be the orthocentre of $\triangle ABC$ .
It follows by simple angle chasing that $\angle ADB = C = 180 - \angle AHB$ . So quadrilateral $AHBD$ is cyclic $\implies \angle AHD = \angle ABD = 180 - 2C$ . Also $\angle AHE = 2C$ . So $\angle AHD + \angle AHE = 180$ which means that $D , E , H$ are collinear as desired.

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548 posts

#8 h Apr 20, 2021, 1:35 AM

Y by

We will use directed angles.

Construct $(BEC)$ , and let $F$ be the intersection of $(BEC)$ and $DE$ . Note that $\measuredangle DFB=\measuredangle ECB=\measuredangle DAB$ , so $F, D, B, A$ are concyclic. I claim that $F$ is the desired orthocenter. To prove this, I will first show that $BF \perp AC$ and then $AF \perp BC$ .

Let $G$ be the intersection of $BF$ and $AC$ . Now, $\measuredangle DEA=\frac{1}{2}\measuredangle DBA=90-\measuredangle BAD$ and $\measuredangle GFE=\measuredangle BCE$ . But, $\measuredangle BAD=\measuredangle BCE$ , so $\measuredangle DEA+\measuredangle GFE=90$ and hence $BF \perp AC$ .

Then, $\measuredangle ACB=\measuredangle DAB=\measuredangle BDF+\measuredangle FDA=\measuredangle BAF+\measuredangle FBA=\measuredangle GFA$ . Since $\measuredangle GFA+\measuredangle FAG=90$ , then $\measuredangle ACB+\measuredangle FAG=90$ and it follows that $AF \perp BC$ . This means that $F$ is the orthocenter of $\triangle ABC$ . Since $F$ lies on $DE$ we're done.

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1184 posts

L567

#9 h Apr 20, 2021, 4:25 AM

Y by

Simple one..

Let $X,Y$ be projections of $B$ onto $AD, AC$ and let $H$ be the orthocenter and $N$ be midpoint of $AH$ . Obviously, $BXAY$ is cyclic.

So, $\angle XYB = \angle XAB = \angle ACB = \angle AHY = \angle NYH$ and so $X,N,Y$ are collinear and so since $X,N,Y$ are midpoints of $AD,AH,AE$ respectively, $D,H,E$ are collinear, as desired

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429 posts

hakN

#10 h Apr 20, 2021, 6:59 AM

Y by

$\angle BHC = 180 - \angle A = 180 - \angle BEA = \angle BEC \implies BHEC$ is cyclic.
Also $\angle DEB = 90 - \frac{1}{2}\angle DBE = 90 - \frac{1}{2}(\angle DBA + \angle ABE) = 90 - \frac{1}{2}(360 - 2\angle C - 2\angle A) = 90 - \angle B$ .
But from $BHEC$ being cyclic, we also know that $\angle HEB = \angle HCB = 90 - \angle B$ .
Thus $\angle DEB = \angle HEB \implies D,H,E$ are collinear. $\square$

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122 posts

#11 h Nov 22, 2022, 1:39 PM • 1 Y

Y by MarioLuigi8972

Bruh what

Complex Bash

This post has been edited 1 time. Last edited by JuanDelPan, Nov 24, 2022, 3:22 AM

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4189 posts

#12 h Aug 11, 2023, 8:43 PM

Y by

Let $\angle A=\alpha$ etc.

Claim: $AHBD$ is cyclic. We have $\angle ADB=\angle DAB=\gamma,$ $\angle AHM=\angle ACB=\gamma.$

This means that $\angle BDE=\angle BED=\angle BAH=90-\beta.$ Thus, if $M$ is the foot from $B$ to $AC$ (also the midpoint of $AE$ ), $\angle HEM=\alpha+\beta-90\rightarrow \angle MHE=\gamma.$ Since $\angle DHB=\angle DAB=\gamma,$ we have $\angle DHB=\angle MHE,$ hence done.

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183 posts

#13 h Dec 5, 2024, 3:47 PM

Y by

Almost direct
Let $DE$ intersect the altitude of $A$ at $H'$
$A,H',B,D$ are concyclic, since $\angle BDE = \angle BDH'= \angle BAH'$ . It implies $H'=H$
We are done

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653 posts

#14 h Apr 25, 2025, 7:09 AM

Y by

Let $A, D, E$ be $a, d, e$ such that they lie on the unit circle, where $B$ is the origin. Let $O,$ the circumcenter of $(ABC),$ be $o.$ We have that $o + \overline o a^2 = a,$ and $o - \overline o ad = a - d.$ Hence, $\overline o = \frac d{a(a + d)} \implies o = \frac{a^2}{a + d}.$ To compute $C,$ we can drop an altitude from $O$ to $\overline{AE}.$ Thus, we get that $c = \frac{a^2 + ea}{a + d}.$ We can now compute the orthocenter by shifting $(ABC)$ to the unit circle through the function $\tau\colon\alpha\to|a + d|(\alpha - \frac{a^2}{a + d}).$ We get that the orthocenter of $\tau(A)\tau(B)\tau(C)$ is $|a + d|(\frac{ea + ad - a^2}{a + d}),$ and hence the orthocenter of $ABC$ is $h = \frac{ea + ad}{a + d}.$ Clearly, $\overline h = \frac{d + e}{(a + d)e}.$ We see that
$\begin{align*} h + ed \overline h = \frac{a(e + d)}{a + d} + \frac{d(e + d)}{a + d} = e + d \end{align*}$ and hence, $E,D, H$ are collinear.

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#15 h Apr 28, 2025, 3:03 AM

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Solution

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