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Let quadrilateral $ABCD$ satisfy $\angle BAC = \angle CAD = 2\,\angle ACD = 40^\circ$ and $\angle ACB = 70^\circ$. Find $\angle ADB$.

What I tried

  1. Ceva’s Theorem (Trigonometry version)
  2. Try to construct some equilateral triangle.

Which both failed.

Any hints or solutions please. Thanks in advance!

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    $\begingroup$ Such upvoting (+6) is exaggerated : the OP hasn't really shown he has worked and hasn't even provided a figure ! $\endgroup$
    – Jean Marie
    Aug 29 '20 at 12:48
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    $\begingroup$ As this question is very easy with the help of some trigonometry, I presume the use of calculator is disallowed? $\endgroup$
    – Amadeus
    Aug 29 '20 at 13:28
  • $\begingroup$ Yes , @l1mbo. Calculator isn’t allowed. $\endgroup$
    – dark.nes_s
    Aug 30 '20 at 1:18
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    $\begingroup$ @JeanMarie, actually the OP has shown how he has worked and the figure is not necessary to understand this question. So he deserves more upvoting than six. $\endgroup$
    – Angelo
    Aug 30 '20 at 8:48
  • $\begingroup$ @Angelo, there are some standards. This is just a posted question and the upvotes probably go to the author of the question. Instead of opposing a more experienced community member, if you had already decided to post an answer to a question, you could've at least drawn your own image. No work is shown in the question, whatsoever. A figure is necessary for others to see what progress the OP made. In fact, (at least) in my country, nobody will give you any points on the state exam without a drawing. $\endgroup$
    – Invisible
    Sep 6 '20 at 16:40
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Well, Geogebra says it is $\approx 77,34^{\circ}$, so good luck... enter image description here

Actually, Ceva might really help:

$${\sin 80\over \sin 40}{\sin(70-x)\over \sin x}{\sin 20\over \sin90} = 1$$

After some manipulation we get $$\cot x = \tan 20+{2\over \cos 10}\implies x =... $$

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  • $\begingroup$ Thanks for all downvotes, care to explain why? $\endgroup$
    – Aqua
    Aug 29 '20 at 14:05
  • $\begingroup$ How did you use ceva , like isn't ceva used in a triangle when 3 cevians are given ? $\endgroup$ Aug 29 '20 at 14:18
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    $\begingroup$ Yes, you have $ABC$ and with respect to $D$. $\endgroup$
    – Aqua
    Aug 29 '20 at 14:20
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    $\begingroup$ oh, yes, I see . Thanks! and also +1 $\endgroup$ Aug 29 '20 at 14:21
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enter image description here

Since $\angle ACB=\angle ABC=70^\circ$, the triangle $ABC$ is isosceles and $\;\overline{AB}=\overline{AC}$.

By applying the law of sines to the triangle $ACD$, we get that:

$\overline{AD}=\overline{AC}\cdot\cfrac{\sin\angle ACD}{\sin\angle ADC}=\overline{AC}\cdot\cfrac{\sin 20^\circ}{\sin 120^\circ}=\cfrac{2\overline{AC}\sin 20^\circ}{\sqrt{3}}\;.$

And, by applying the law of sines to the triangle $ABD$, we get that:

$\overline{AD}\sin\angle ADB=\overline{AB}\sin\angle ABD\;.\quad\color{blue}{(*)}$

Let $\;\alpha=\angle ADB\;.$

Since $\;\overline{AD}=\cfrac{2\overline{AC}\sin 20^\circ}{\sqrt{3}}\;$, $\;\overline{AB}=\overline{AC}\;$ and $\;\angle ABD=100^\circ-\alpha\;,\;$ the equality $(*)$ turns into:

$\cfrac{2\overline{AC}\sin 20^\circ\sin\alpha}{\sqrt{3}}=\overline{AC}\sin(100^\circ-\alpha)\;,$

$2\sin 20^\circ\sin\alpha=\sqrt{3}\sin(90^\circ+10^\circ-\alpha)\;,$

$4\sin 10^\circ\cos 10^\circ\sin\alpha=\sqrt{3}\cos(10^\circ-\alpha)\;,$

$4\sin 10^\circ\cos 10^\circ\sin\alpha=\sqrt{3}\left(\cos10^\circ\cos\alpha+\sin 10^\circ\sin\alpha\right)\;,$

$4\sin 10^\circ\sin\alpha=\sqrt{3}\left(\cos\alpha+\tan 10^\circ\sin\alpha\right)\;,$

$\left(4\sin 10^\circ-\sqrt{3}\tan 10^\circ\right)\sin\alpha=\sqrt{3}\cos\alpha\;,$

$\tan\alpha=\cfrac{\sqrt{3}}{4\sin 10^\circ-\sqrt{3}\tan 10^\circ}\;.$

Hence,

$\angle ADB=\alpha=\arctan\left(\cfrac{\sqrt{3}}{4\sin 10^\circ-\sqrt{3}\tan 10^\circ}\right)\simeq\\\simeq 77,3361794^\circ.$

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    $\begingroup$ He he, it is like my picture. $\endgroup$
    – Aqua
    Aug 29 '20 at 18:23
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    $\begingroup$ You are right, I do not have a program to draw, so I have used your drawing. I am sorry. $\endgroup$
    – Angelo
    Aug 29 '20 at 22:01
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    $\begingroup$ If you don't have a program, but want to post an answer, you can take a picture of a real drawing. 😊 $\endgroup$
    – Invisible
    Aug 30 '20 at 4:46
  • $\begingroup$ But I am not good at drawing. $\endgroup$
    – Angelo
    Aug 30 '20 at 6:49

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