angle chasing for juniors, equilateral and square on the same side of a line

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angle chasing for juniors, equilateral and square on the same side of a line

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Source: V.A. Yasinsky Geometry Olympiad 2018 VIII-IX p1, advanced p1 [Ukraine]

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#1 h Oct 1, 2018, 6:39 PM • 2 Y

Y by Adventure10, Mango247

Points $A, B$ and $C$ lie on the same line so that $CA = AB$ . Square $ABDE$ and the equilateral triangle $CFA$ , are constructed on the same side of line $CB$ . Find the acute angle between straight lines $CE$ and $BF$ .

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#2 h Oct 1, 2018, 7:33 PM • 2 Y

Y by Adventure10, Mango247

The answer is $75$ . :
Let $K=(AFE) \cap BC$ .We have $\angle{FAE} =30$ . We know that $AE=AF$ $\Longrightarrow \angle{AEF} =\angle{AFE} =75 \Longrightarrow \angle{FKC} =\angle{EKB} =75$ .
Now let $T=(KEB) \cap (FKC)$ . It is enough to prove that $B, T$ and $F$ are collinear (in the same way we can prove that $C, T$ and $E$ are collinear).
We have :
$\angle{KTB} =\angle{KEB} =60=180-\angle{FTK}$ As desired...
$\blacksquare$

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#3 h Nov 9, 2018, 6:01 PM • 2 Y

Y by Adventure10, Mango247

Since, $AC=EA \implies \Delta CFA \text{ is isosceles-right triangle } \implies \boxed{\angle ECA=45^{\circ}}$ ,
If $AB=x$ and Let $FT \perp CB$ such, $T \in CB$ , Hence, $FT=\frac{x\sqrt{3}}{2} \text{ and } TB=\frac{3x}{2} \implies \tan \angle CBF =\frac{\tfrac{x\sqrt{3}}{2}}{\tfrac{3x}{2}}=\frac{1}{\sqrt{3}} \implies \boxed{\angle CBF=30^{\circ}}$
Therefore, the acute angle between the lines $CE$ and $FB$ will be $75^{\circ}$

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zuss77

#4 h Jun 12, 2020, 11:58 AM

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Extremely trivial. $\angle ACE=45^\circ$ , $AB=AC=AF \implies \angle BFC=90^\circ, \angle CBF=30^\circ, \angle ? = 75^\circ$ .

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jayme

#5 h Jul 13, 2020, 11:06 AM • 1 Y

Y by Mango247

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20X.pdf p. 38...

Sincerely
Jean-Louis

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#6 h Jul 13, 2020, 11:50 AM • 1 Y

Y by Mango247

Observe that $BC=BD$ . Then, we have $\angle BCD=45^\circ$ . Observe that $AB=BF$ . We have
$\angle AFB=\frac{180^\circ - \angle ABF}{2} = \frac{\angle CBF}{2}=30^\circ$ Then, $\angle HFC=90^\circ$ . Because $\angle BCF=60^\circ$ , we have
$\angle DCF=\angle BCF-\angle BCD=60^\circ-45^\circ=15^\circ$ And then
$\angle FGC=180^\circ-\angle GFC-\angle GCF=180^\circ-90^\circ -15^\circ = \boxed{75^\circ}$

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