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 $\angle BHF=90^\circ$.
Problem
Source: IGO 2021 Advanced P1
Tags: geometry, circumcircle, IGO
30.12.2021 19:45
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.
30.12.2021 21:02
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][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][/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 $\angle DEC = \angle A'BC = \angle C$. $\blacksquare$
08.01.2022 15:09
08.01.2022 15:55
Proposed by Harris Leung - Hong Kong
10.01.2022 10:11
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 $\triangle BHF\sim\triangle A'AE$.
15.01.2022 19:28
$B'$ 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. $\blacksquare$
25.01.2022 22:52
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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17.04.2022 11:22
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.
20.04.2022 20:31
$\angle HBC+\angle BAH=\angle A$, So, $\angle AHB=\angle C$, Thus, by obvious chase $AEHF$ is cyclic.So, done. Note:- I didn't cheat above solutions.
25.07.2022 09:51
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 $H$.
17.12.2022 07:45
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, $\angle BHF = \angle AHF - \angle BHA = \angle AEF - \angle ACB = 90^{\circ} + \angle BEF - \angle ACB \stackrel{DE=DC}{=} = 90^{\circ}$ $\blacksquare$
11.04.2023 15:14
30.09.2023 17:24
I know it is to some above but just for storage
20.10.2023 13:57
$\angle FEH=\angle BEH-\angle BEF=\angle ACH-\angle DEC=\angle ACH-\angle ACB=\angle BCH=\angle FAH$ 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.
20.10.2023 14:02
This point H is also point X in ISL2011G4. I thought I could use that problem, but it was superfluous lol.
20.04.2024 18:35
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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