angles of trapezoid ABCD wanted, AD = FB, BC = AF, < BEC = 54^o

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angles of trapezoid ABCD wanted, AD = FB, BC = AF, < BEC = 54^o

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Source: (2021-) 2022 XV Dürer Math Competition Finals Day1 D4

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#1 h Dec 1, 2022, 7:29 PM

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The longer base of trapezoid $ABCD$ is $AB$ , while the shorter base is $CD$ . Diagonal $AC$ bisects the interior angle at $A$ . The interior bisector at $B$ meets diagonal $AC$ at $E$ . Line $DE$ meets segment $AB$ at $F$ . Suppose that $AD = FB$ and $BC = AF$ . Find the interior angles of quadrilateral $ABCD$ , if we know that $\angle BEC = 54^o$ .

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#2 h Dec 2, 2022, 8:21 PM • 1 Y

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$\alpha=72^{\circ},\beta=36^{\circ},\gamma=144^{\circ},\delta=108^{\circ}$ .

This post has been edited 1 time. Last edited by vanstraelen, Dec 3, 2022, 9:07 AM
Reason: overwriting error

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#3 h Dec 3, 2022, 6:30 AM

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vanstraelen wrote:

$\alpha=72^{\circ},\beta=36^{\circ},\gamma=144^{\circ},\delta=72^{\circ}$ .

$\delta=108^\circ$

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#4 h Dec 3, 2022, 3:39 PM

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Let $\angle CAB=a$ and $\angle EBA=b$ .
Rule of sines in $\triangle ABC\ :\ \frac{AC}{\sin 2b}=\frac{AB}{\sin(54^{\circ}+b)}=\frac{BC}{\sin a}$ .
Rule of sines in $\triangle ACD\ :\ \frac{AD}{\sin(54^{\circ}-b)}=\frac{AC}{\sin 2a}$ .
Angle chasing: $a+b=54^{\circ}$ .

$\frac{AF+FB}{\sin(54^{\circ}+b)}=\frac{BC}{\sin a}$ ,
$\frac{BC+AD}{\sin(54^{\circ}+b)}=\frac{BC}{\sin a}$ ,
$BC[\sin(54^{\circ}+b)-\sin a]=AD\sin a$ ,
$\frac{AC\sin a}{\sin 2b} \cdot [\sin(54^{\circ}+b)-\sin a]=\frac{AC\sin(54^{\circ}-b)}{\sin 2a} \cdot \sin a$ ,

$\sin 2a \cdot [\sin(54^{\circ}+b)-\sin a]=\sin(54^{\circ}-b)\sin 2b$ ,
$\sin(108^{\circ}-2b) \cdot [\sin(54^{\circ}+b)-\sin(54^{\circ}-b)]=\sin(54^{\circ}-b)\sin 2b$ ,
$\sin(72^{\circ}+2b) \cdot 2\sin b\cos 54^{\circ}=\sin(54^{\circ}-b) \cdot 2\sin b\cos b$ ,
$2\sin(72^{\circ}+2b)\cos 54^{\circ}=2\sin(54^{\circ}-b)\cos b$ ,

$\sin(126^{\circ}+2b)+\sin(18^{\circ}+2b)=\sin 54^{\circ}+\sin(54^{\circ}-2b)$ ,
$\sin(18^{\circ}+2b)=\sin 54^{\circ}$ ,
$18^{\circ}+2b=54^{\circ}$ and $b=18^{\circ}$ .
$\beta=2b=36^{\circ}$ .

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