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Source: IGO 2015 Intermediate 2

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#1 h Jul 16, 2018, 11:42 PM • 4 Y

Y by Adventure10, Mango247, Rounak_iitr, DEKT

In acute-angled triangle $ABC$ , $BH$ is the altitude of the vertex $B$ . The points $D$ and $E$ are midpoints of $AB$ and $AC$ respectively. Suppose that $F$ be the reflection of $H$ with respect to $ED$ . Prove that the line $BF$ passes through circumcenter of $ABC$ .

by Davood Vakili

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#2 h Jul 17, 2018, 12:42 AM • 3 Y

Y by David2663, Adventure10, Mango247

Let $K$ be on line $DE$ such that $BK$ is the internal angle bisector of $\angle ABC$ . Note that $DA=DB=DK$ by angle-chasing, so thus $H,K$ lie on the circle with diameter $AB.$ $F$ also lies on this circle since it's the reflection of $H$ over a line through the center, $D$ . Hence, $BHKF$ is cyclic, while $KF=KH$ implies that $BK$ also bisects $\angle FBH$ . Hence, $BF$ is isogonal to $BH$ and passes through $O$ .

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#3 h Feb 28, 2019, 1:37 PM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
Solution
Let $S$ be midpoint of $BC$ ; $K$ $\equiv$ $BH$ $\cap$ $DE$
We have: ( $DB$ ; $DF$ ) $\equiv$ ( $DB$ ; $DE$ ) + ( $DE$ ; $DF$ ) $\equiv$ ( $BA$ ; $BC$ ) + ( $DH$ ; $DE$ ) $\equiv$ ( $BA$ ; $BC$ ) + ( $DH$ ; $DS$ ) + ( $DS$ ; $DE$ ) $\equiv$ ( $BA$ ; $BC$ ) + ( $SD$ ; $SE$ ) + ( $CA$ ; $CB$ ) $\equiv$ ( $BA$ ; $BC$ ) + ( $AC$ ; $AB$ ) + ( $CA$ ; $CB$ ) $\equiv$ 2 ( $CA$ ; $CB$ ) $\equiv$ 2 ( $EA$ ; $ED$ ) $\equiv$ ( $EA$ ; $EF$ ) $\equiv$ ( $EA$ ; $EF$ ) $\equiv$ ( $KB$ ; $KF$ ) (mod $\pi$ )
So: $B$ , $D$ , $K$ , $F$ lie on a circle
Then: ( $BF$ ; $BD$ ) $\equiv$ ( $KF$ ; $KE$ ) $\equiv$ ( $KE$ ; $KH$ ) $\equiv$ ( $BC$ ; $BH$ ) (mod $\pi$ ) or $BF$ , $BH$ are isogonal conjugate with respect to $\triangle$ $ABC$
Hence: $BF$ passes through circumcenter of $\triangle$ $ABC$

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#4 h Aug 11, 2019, 8:58 PM • 3 Y

Y by BVKRB-, Adventure10, Mango247

It can be done using Complex bash also, Set $\odot(ABC)$ as the unit circle, so, $h=\frac{1}{2}(a+b+c-\frac{ac}{b})$ , also $d=\frac{a+b}{2}$ and $e=\frac{a+c}{2}$ . Now by the reflection lemma you get $f$ . Now we see that $\frac{b-f}{b-o}\in\mathbb R$ , where $o=0$ . So, $B,F,O$ are collinear.

This post has been edited 2 times. Last edited by amar_04, Aug 11, 2019, 8:58 PM

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#5 h Aug 11, 2020, 5:06 PM • 2 Y

Y by Rounak_iitr, Siddharthmaybe

The easiest solution (only angle chase).

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Mahdi_Mashayekhi

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#6 h Dec 26, 2021, 7:07 PM • 1 Y

Y by IceWolf10

We only need to prove ∠FBA = 90 - ∠C.
∠AHB = 90 ---> DA = DH = DB
F is refection of H about DE ---> DH = DF
so DA = DH = DB = DF and it means AHFB is cyclic so ∠FBA = ∠FHE which is 90 - ∠c.
we're Done.

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#7 h Aug 16, 2023, 5:00 PM

Y by

Interesting problem
Here is my solution:
We extend $BF$ which intersects the circumcircle of $\bigtriangleup ABC$ at $I$ .Then join $IC,EF,DF,DH$
Claim 1 : $BAHF$ is cyclic

Proof : In $\bigtriangleup ABH$ , We have $DA=DB=DH$ .Since $DE$ is the perpendicular bisector of $HF$ , $DHEF$ is a kite.
Thus, $DH=DF=DA=DB$ implies that indeed quadrilateral $BAHF$ is cyclic with center $D$ .

Claim 2 : $\angle BCI=90^{\circ}$

Proof : Extend $HF$ which intersects $BC$ at $G$ .Then we have $\angle HBG=\angle GHC\equiv \angle FHC=\angle ABF$ .
Then $\angle ABH=\angle ABF-\angle HBF=\angle HBG-\angle HBF=\angle FBG\equiv \angle IBC$
Now $\angle ABH=\angle IBC$ and $\angle BAC\equiv \angle BAH=\angle BIC$ .Thus $\bigtriangleup ABH\sim\bigtriangleup IBC$ implies that $\angle BCI=\angle BHA=90^{\circ}$

So, from claim 2 we conclude that $BI$ is the diameter of $(ABC)$ and $BF$ passes through the circumcenter of $\bigtriangleup ABC$ $\square$

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This is my first ever post with LATEX . So feel free to spot any error .Thank you

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#8 h Aug 16, 2023, 7:56 PM

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Let $\alpha, \beta, \gamma$ be the angles $\angle A, \angle B, \angle C,$ respectively. Since $D$ is the circumcenter of $AHB,$ we must have that $DA = DF = DH = DB.$ In particular, $DF = DB.$

Now, we have that $\angle ADF = \angle ADE + \angle EDF = 2 \angle ADE - \angle ADH = 2 \beta - (180 - 2 \alpha) = 180 - 2 \gamma.$ This means that $\angle BDF = 2 \gamma.$ Since $DF = DB,$ we conclude that $\angle ABF = \angle DBF = 90^\circ - \gamma.$ But $\angle CBH = 90^\circ - \gamma,$ so the lines $BH$ and $BF$ are isogonal. It is well-known that the $B$ -altitude is isogonal to the $B$ -diameter line, so $BF$ passes through the circumcenter of $\triangle ABC.$ Hence done.

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