Proving &ang;BHF=90

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Proving ∠BHF=90

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Source: IGO 2021 Advanced P1

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577 posts

#1 h Dec 30, 2021, 4:45 PM • 3 Y

Y by Rounak_iitr, buddyram, ItsBesi

Acute-angled triangle $ABC$ with circumcircle $\omega$ is given. Let $D$ be the midpoint of $AC$ , $E$ be the foot of altitude from $A$ to $BC$ , and $F$ be the intersection point of $AB$ and $DE$ . Point $H$ lies on the arc $BC$ of $\omega$ (the one that does not contain $A$ ) such that $\angle BHE=\angle ABC$ . Prove that $\angle BHF=90^\circ$ .

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577 posts

#2 h Dec 30, 2021, 4:45 PM • 3 Y

Y by teomihai, miko286k, Rounak_iitr

Let $HE\cap \omega=\{H, K\}$ .
We have $\angle ABC=\angle BHE=\angle BHK=\angle BCK\Rightarrow ABCK$ is an isosceles trapezoid.
Hence, $AK//BC\Rightarrow \angle EAK=90^\circ$ .
Let $L$ be the symmetry of $E$ wrt $D$ .
We know that $AECL$ is a rectangle. Hence, $\angle EAL=90^\circ=\angle EAK\Rightarrow A, K, L$ are collinear.
Let $LC\cap \omega=\{C,T\}$ .

Apply Pascal to $KABCTH$ .
We have $KA\cap CT=L, BC\cap KH=E$ . Hence, $AB\cap TH$ lies on $LE$ . Also, $AB\cap LE=F$ . Thus, $F$ lies on $TH$ . Then, $\angle BHT=\angle BCT=\angle ECL=90^\circ\Rightarrow \angle BHF=90^\circ$ , as desired.

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656 posts

#3 h Dec 30, 2021, 6:02 PM

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Let $L = \overline{AE} \cap \omega \ne A, A' = \overline{HE} \cap \omega \ne H$ and $B_0$ be the antipode of $B$ wrt $\omega$ . Our problem is equivalent to showing that points $F,H,B_0$ are collinear.
$[asy] size(200); pair A = dir(120),B=dir(-155),C=dir(-25),O=(0,0); draw(unitcircle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); pair Ap = dir(60), E = foot(A,B,C), D = 1/2*(A+C), F = extension(A,B,D,E),H = IP(E--2*E-Ap,unitcircle); dot("$A'$",Ap,dir(Ap)); dot("$H$",H,dir(-90)); dot("$E$",E,dir(130)); dot("$D$",D,dir(D)); dot("$F$",F,dir(F)); pair L = 2*E - A - B - C, B0 = -B; dot("$L$",L,dir(20)); dot("$B_0$",B0,dir(B0)); draw(B--C^^A--Ap,red); draw(C--A--F,royalblue); draw(B--B0^^L--Ap,purple); draw(B--Ap^^F--D,green); draw(A--L^^H--Ap,brown); draw(F--B0,dashed); [/asy]$
Note that $A'$ is the reflection of $A$ in the perpendicular bisector of segment $BC$ and so $A'L$ is a diameter of $\omega$ , in particular $\overline{BA'} \parallel \overline{B_0L}$ . Now Pascal on $BALB_0HA'$ gives that our problem is equivalent to $\overline{DE} \parallel \overline{BA'}$ . But that is clear as $D$ is the circumcenter of $\triangle AEC$ which gives $\angle DEC = \angle A'BC = \angle C$ . $\blacksquare$

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256 posts

#5 h Jan 8, 2022, 12:09 PM • 4 Y

Y by teomihai, Mehrshad, Sondat_theorem, Rounak_iitr

Solution

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181 posts

#6 h Jan 8, 2022, 12:55 PM

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Proposed by Harris Leung - Hong Kong

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361 posts

#7 h Jan 10, 2022, 7:11 AM

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Let $A'\in\omega$ be such that $AA'\parallel BC$ .

1- trivially $A',E,H$ are collinear.
2- with Menelaus you get $BF=\frac{c\cdot BE}{EC-BE}=\frac{c\cdot BE}{AA'}$ .
3- $\frac{BE}{A'E}=\frac{BH}{A'C}=\frac{BH}{c}\Rightarrow BH=\frac{c\cdot BE}{A'E}$
4- angle chasing to show $\angle FBH=\angle AA'E$ .
5- use everything to show $\triangle BHF\sim\triangle A'AE$ .

This post has been edited 1 time. Last edited by hectorraul, Jan 10, 2022, 7:23 AM

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#8 h Jan 15, 2022, 4:28 PM

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$B'$ is the antipode of $B$ wrt $(ABC)$ . Define $H=FB' \cap (ABC)$ . It is enough to prove that $\angle BHE=\angle ABC$ .

$\textbf{Claim:}$ $HBE \sim HAC$ .
Proof: $\angle HBE= \angle HAC$ and $\frac{HB}{HA}\overset{\text{ratio lemma}}=\frac{FB}{FA} \cdot \frac{B'A}{B'B}\overset{\text{Menelaus}}=\frac{CD}{AD} \cdot \frac{EB}{EC} \cdot \frac{B'A}{B'B}=EB \cdot \frac{B'A}{CE \cdot B'B}\overset{BAB' \sim AEC}=\frac{EB}{AC}$ as needed. $\square$

Thus $\angle BHE=\angle AHC=\angle ABC$ as desired. $\blacksquare$

This post has been edited 2 times. Last edited by SerdarBozdag, Jan 15, 2022, 4:41 PM

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Neo-Pythagorean

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Neo-Pythagorean

#9 h Jan 25, 2022, 7:52 PM

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Let $H'$ be the second intersection of the circumcircles of $\triangle{ABC}$ and $\triangle{AEF}$ . We have that $\angle{AH'F}=\angle{AEF}=90^\circ+\angle{C}$ and $\angle{AH'B}=\angle{C}$ , so $\angle{BH'F}=\angle{AH'F}-\angle{AH'B}=90^\circ$ . So, it suffices to show that $H'\equiv{H} \Longleftrightarrow \angle{BH'E}=\angle{ABC}$ . We have $\angle{AH'E}=\angle{AFE}=\angle{AED}-\angle{BAE}=(90^\circ-\angle{C})-(90^\circ-\angle{B})=\angle{B}-\angle{C} \Longrightarrow \angle{BH'E}=\angle{BH'A}+\angle{AH'E}=\angle{ABC}$ . The proof is complete.

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930 posts

#10 h Apr 17, 2022, 8:22 AM • 1 Y

Y by Rounak_iitr

Three liner solution :
$\angle EHC=\angle BHC-\angle BHE=180^o-A-B=C=\angle DCE=\angle DEC$ Thus from condition of tangency we have $(HEC)$ is tangent to the line $DF$ at $E$ .
Thus,
$\angle FAH=\angle BAH=\angle BCH=\angle FEH \implies AEHF \text{is cyclic.}$ Thus $\angle BHF=\angle AHF-\angle AHB=\angle AEF-\angle ACB=90^o+C-C=90^o$ as desired.

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VJ22

#11 h Apr 20, 2022, 5:31 PM

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$\angle HBC+\angle BAH=\angle A$ , So, $\angle AHB=\angle C$ , Thus, by obvious chase $AEHF$ is cyclic.So, done.
Note:- I didn't cheat above solutions.

This post has been edited 1 time. Last edited by VJ22, Apr 20, 2022, 5:32 PM
Reason: .

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#12 h Jul 25, 2022, 6:51 AM

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Let $AEF$ meet $ABC$ at $H'$ .
Claim $: \angle BH'F = \angle 90$ .
Proof $:$ Note that $\angle AH'F = \angle AEF = \angle 90 + \angle BEF = \angle 90 + \angle DEC = \angle 90 + \angle DCE = \angle 90 + \angle ACB = \angle 90 + \angle AH'B \implies BH'F = \angle 90$ .
Claim $: \angle BH'E = \angle ABC$
Proof $:$ Note that $\angle BEH' = \angle BEF + \angle FEH' = \angle ACB + \angle FAH' = \angle ACB + \angle BAH' = \angle ACB + \angle BCH' = \angle ACH' = \angle FBH' \implies FB$ is tangent to $BEH' \implies \angle BH'E = ABC$ so $H'$ is $H$ .

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Jul 25, 2022, 6:52 AM

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#13 h Dec 17, 2022, 4:45 AM

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Claim: $AEHF$ is cyclic.
Proof: It's clear that $D$ is circumcenter of $\triangle AEC$ , then $DA = DE = DC$ . Notice that $\angle AFE = \angle AFD = 180^{\circ} - \angle FAD -\angle FDA = 180^{\circ} - \angle BAC - 2\angle ACB$ .
Now $\angle BHE = \angle ABC = AHC$ $\Rightarrow$ $\angle CHE = \angle ACB = \angle AHB$ then $\angle AHE = \angle BHC - \angle AHB - \angle CHE= 180^{\circ} - \angle BAC - 2\angle ACB = \angle AFE$ $\square$
Finally, $\angle BHF = \angle AHF - \angle BHA = \angle AEF - \angle ACB = 90^{\circ} + \angle BEF - \angle ACB \stackrel{DE=DC}{=} = 90^{\circ}$ $\blacksquare$

This post has been edited 1 time. Last edited by UI_MathZ_25, Dec 17, 2022, 4:46 AM

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#14 h Apr 11, 2023, 12:14 PM • 1 Y

Y by Rounak_iitr

Solution 1 (Angle chasing)
Solution 2 (√bc/2 inversion)

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#15 h Sep 30, 2023, 2:24 PM

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I know it is to some above but just for storage
storage

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#16 h Oct 20, 2023, 10:57 AM

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$\angle FEH=\angle BEH-\angle BEF=\angle ACH-\angle DEC=\angle ACH-\angle ACB=\angle BCH=\angle FAH$ so we get $A,E,H,F$ are concyclic. Thus, $\angle BHF=180^\circ-\angle BAE-\angle BHE=180^\circ-(90^\circ-\angle ABC)-\angle ABC=90^\circ$ as desired.

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#17 h Oct 20, 2023, 11:02 AM

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This point H is also point X in ISL2011G4. I thought I could use that problem, but it was superfluous lol.

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884 posts

#18 h Apr 20, 2024, 3:35 PM

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Let the point $A'$ be the reflection of $A$ about the perpendicular bisector of $BC$ . Let $B'$ be the antipode of $B$ w.r.t $w$ . Let $CB'$ and $A'A$ intersect at $X$ . Let $B'H$ intersect $AB$ at $F'$ . Then applying Pascals Theorem on $BCB'HA'A$ . We get that the point $X$ , $E$ , and $F'$ are collinear. Since $X$ lies on $ED$ ( $AECX$ is a rectangle) we must have $F=F'$ . Now $\angle BHF =\angle BHB'=90^{\circ}$ .

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#19 h Apr 2, 2025, 5:38 AM

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Claim 1: $AEHF$ is cyclic.
Notice that $\angle FBE+\angle EFB=\angle FEC \iff \angle ABC+\angle EFA=\angle FEC$ , In circumcircle of $\triangle AEC$ with $D$ midpoint of diameter $AC$ , We obtain that $\angle FEC=\angle DEC=\angle ECD$ , also we have $\angle ECD=\angle ECA=\angle BHA=\angle BHE+\angle EHA$ by $ABHC$ cyclic. then $\angle BHE+\angle EHA=\angle ABC+\angle EFA \iff \angle EHA=\angle EFA$ that means $AEHF$ is cyclic.

We know that $\angle FAE=180^\circ-\angle EAB \iff 180^\circ=\angle FAE+\angle EAB$ . By Claim 1, we obtain that $180^\circ=\angle FAE+\angle EHF$ , then $\angle FAE+\angle EAB=\angle FAE+\angle EHF \iff \angle EAB=\angle EHF$ . Notice in $\triangle EAB$ that $\angle EAB=90^\circ-\angle ABE\iff \angle EAB=90^\circ-\angle ABC$ and then $\angle EHF=90^\circ-\angle ABC \iff \angle EHF+\angle ABC=90^\circ$ . By hypothesis, we obtain that $\angle EHF+\angle ABC=90^\circ \iff \angle EHF+\angle BHE=90^\circ \iff \angle BHF=90^\circ$ and we are done.

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This post has been edited 2 times. Last edited by jordiejoh, Apr 2, 2025, 5:39 AM

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