angles given, trapezoid wanted (2009 Zhautykov City IX P4)

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angles given, trapezoid wanted (2009 Zhautykov City IX P4)

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#1 h Sep 5, 2019, 6:30 AM • 1 Y

Y by Adventure10

In a convex quadrangle $ABCD$ , $\angle BAC = \angle DBC = 30 {} ^\circ, \angle BCA = 20 {} ^\circ$ and $\angle BDC = 70 {} ^\circ$ . Prove that $ABCD$ is a trapezoid.

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#2 h Sep 8, 2019, 3:50 PM • 1 Y

Y by Adventure10

Rule of sines, three times; $AB=1$ .

$\triangle ABC\ :\ \frac{1}{\sin 20^{\circ}}=\frac{AC}{\sin 50^{\circ}}=\frac{BC}{\sin 30^{\circ}}$ .
$\triangle BCD\ :\ \frac{CD}{\sin 30^{\circ}}=\frac{BC}{\sin 70^{\circ}}$ .
$\triangle ACD\ :\ \frac{CD}{\sin x}=\frac{AC}{\sin(120^{\circ}-x)}$ .

$\Rightarrow\ \frac{\sin 30^{\circ}\sin 30^{\circ}}{\sin x \sin 70^{\circ}\sin 20^{\circ}}=\frac{\sin 50^{\circ}}{\sin 20\sin(120^{\circ}-x)}$ .
$4\sin 50^{\circ}\sin 70^{\circ}\sin x=\sin(120^{\circ}-x)$ .
$2(\cos 20^{\circ}-\cos 120^{\circ})\sin x=\sin(120^{\circ}-x)$ .

$(2\cos 20^{\circ}+1)\sin x=\sin 120^{\circ}\cos x-\cos 120^{\circ}\sin x$ .
$(2\cos 20^{\circ}+1-\frac{1}{2})\sin x=\sin 60^{\circ}\cos x$ .
$(2\cos 20^{\circ}+\frac{1}{2})\sin x=\sin 60^{\circ}\cos x$ .

$(\cos 20^{\circ}+\cos 20^{\circ}+\cos 60^{\circ})\sin x=\sin 60^{\circ}\cos x$ .
$(\cos 20^{\circ}+2\cos 40^{\circ}\cos 20^{\circ})\sin x=(3\sin 20^{\circ}-4\sin^{3}20^{\circ})\cos x$ .
$\cos 20^{\circ}(1+2\cos 40^{\circ})\sin x=\sin 20^{\circ}(3-4\sin^{2}20^{\circ})\cos x$ .

$\cos 20^{\circ}(1+4\cos^{2}20^{\circ}-2)\sin x=\sin 20^{\circ}(3-4\sin^{2}20^{\circ})\cos x$ .
$\cos 20^{\circ}\sin x=\sin 20^{\circ}\cos x$ .
$x=20^{\circ}$ , so $\angle CAD =20^{\circ}$ and $\angle ABD=30^{\circ}=\angle DBC \Rightarrow AD // BC$ .

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