DC + DL <= BC wanted , <C > 60^o, CA=CB, angle bisector, circumcircle related

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DC + DL <= BC wanted , <C > 60^o, CA=CB, angle bisector, circumcircle related

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#1 h Jul 17, 2020, 9:46 PM

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Given the isosceles triangle $ABC$ ( $CA = CB$ ). The bisector of the angle $\angle B$ intersects the side $AC$ at point $L$ and the circumcircle of the triangle $ABC$ at point $D$ . It is known that $\angle C > 60^o$ . Prove that $DC + DL \le BC$ .

(I.Voronovich)

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#2 h Aug 11, 2020, 9:05 AM • 1 Y

Y by Mango247

Let $\angle A=2\theta$ . Then $\theta\in(0^\circ, 30\circ)$ and we need to prove that
$DC+DL\le BC$ $\iff$
$DC+\frac{DC^2}{DB}\le BC$ $\iff$
$\sin\theta+\frac{\sin^2\theta}{\sin 3\theta}\le \sin 2\theta$ $\iff$
$\sin\theta+\sin 3\theta\le 2\cos\theta\sin 3\theta$ $\iff$
$\sin 2\theta\le \sin 3\theta$ which is obvious.

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