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

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.

• Such upvoting (+6) is exaggerated : the OP hasn't really shown he has worked and hasn't even provided a figure ! Aug 29 '20 at 12:48
• As this question is very easy with the help of some trigonometry, I presume the use of calculator is disallowed? Aug 29 '20 at 13:28
• Yes , @l1mbo. Calculator isn’t allowed. Aug 30 '20 at 1:18
• @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. Aug 30 '20 at 8:48
• @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. Sep 6 '20 at 16:40

Well, Geogebra says it is $$\approx 77,34^{\circ}$$, so good luck...

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 $$\cot x = \tan 20+{2\over \cos 10}\implies x =...$$

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

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,

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

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