prove equal angles starting with an obtuse triangle

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prove equal angles starting with an obtuse triangle

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Source: Mexican Mathematical Olympiad 2000 OMM P6

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#1 h Jul 28, 2018, 10:04 AM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Let $ABC$ be a triangle with $\angle B > 90^o$ such that there is a point $H$ on side $AC$ with $AH = BH$ and BH perpendicular to $BC$ . Let $D$ and $E$ be the midpoints of $AB$ and $BC$ respectively. A line through $H$ parallel to $AB$ cuts $DE$ at $F$ . Prove that $\angle BCF = \angle ACD$ .

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#2 h Jul 28, 2018, 1:17 PM • 2 Y

Y by Adventure10, Mango247

Let $G=BF \cap AC$ .

From $H$ on the perpendicular bisector of $AB$ which passes through $D$ we get by angle chasing that $BFHD$ is cyclic. Furthermore we get:
$\angle GBA=\angle FBD=90^{\circ}$ Combining this with $\angle B=\angle A+90^{\circ}$ shows $CB$ is tangent to $\odot GBA$ giving:
$CB^2=CG \cdot CA$
From $F$ is on the $B$ -midline we get $BF=FG$ so:
$\frac{\sin{\angle FCB}}{\sin{\angle ACF}}=\frac{CG}{CB}=\frac{CB}{CA}$ Which is enough to show $F$ lies on the $C$ -symmedian in $\triangle ABC$ as desired.

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#3 h Jul 28, 2018, 2:35 PM • 2 Y

Y by Adventure10, Mango247

To make little bit more attractive the previous solution: $DAHF$ is a parallelogram, $HF=AD=BD$ and thus $BDHF$ is a rectangle and $\angle CBF=\angle BAC$ , that is, $AB$ and $BF$ are antiparallel; $CF$ being median for $\triangle BCF$ , it is symmedian for $\triangle ABC$ , done!

Best regards,
sunken rock

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#4 h Nov 26, 2022, 12:13 AM • 1 Y

Y by Krave37

Very nice problem.

Solution

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#5 h Oct 18, 2024, 7:05 AM

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As $AC$ and $DE$ are parallel, $\angle ACD =\angle CDE$ , Now draw a line parallel to $CF$ passing through $B$ and extend $DE$ to hit the line parallel to $CF$ at $O$ , This gives $\angle BCF = \angle CBO$ , connecting $OC$ , we are left with proving $BOCD$ is cyclic, now we have to prove $\angle BDO = \angle BCO$ , $BF$ is parallel to $OC$ as $E$ is the midpoint of $BC$ , so $\angle BCO= \angle CBF$
As we now have to show $\angle CBF = \angle BDO= \angle BDF$ , we have to show that circumcenter of $BFD$ lies on $BH$ so $BE$ is a tangent, as $AHB$ is isosceles, $\angle HAB= \angle HBA$ , if $BH$ intersects $ED$ at $Z$ ,
$ZD=ZB$ , now we have to show $ZD=ZB=ZF$ for $Z$ to be the circumcenter, by angle chasing we get $\angle BHF = \angle BAH$ and $\angle BZO= \angle BHC = \angle 2BAH$ , so for triangle $BHF$ , $Z$ is the circumcenter, giving $ZB=ZF$ .

This post has been edited 1 time. Last edited by Krave37, Oct 26, 2024, 7:52 PM

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#6 h Oct 26, 2024, 7:20 PM

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sunken rock wrote:

To make little bit more attractive the previous solution: $DAHF$ is a parallelogram, $HF=AD=BD$ and thus $BDHF$ is a rectangle and $\angle CBF=\angle BAC$ , that is, $AB$ and $BF$ are antiparallel; $CF$ being median for $\triangle BCF$ , it is symmedian for $\triangle ABC$ , done!

Best regards,
sunken rock

Can you clarify, how is CF a median of BCF. Isnt it one of its sides? you meant BCG?

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#7 h Oct 27, 2024, 2:51 AM

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$ADFH$ is a parallelogram by midline and given conditions. Now $BD=AD=HF$ so $HDBF$ parallelogram.

first isogonality lemma on $HDBF$ finishes.

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#8 h Oct 27, 2024, 7:53 AM

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TestX01 wrote:

$ADFH$ is a parallelogram by midline and given conditions. Now $BD=AD=HF$ so $HDBF$ parallelogram.

first isogonality lemma on $HDBF$ finishes.

whats MIDLINE and Isogonality Lemma

This post has been edited 1 time. Last edited by Krave37, Oct 27, 2024, 7:53 AM

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#9 h Oct 28, 2024, 6:51 AM

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check out lemmas in aops geometry https://web.evanchen.cc/handouts/GeoSlang/GeoSlang.pdf. midline is line joining midpoints

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