# Geometry problem, finding missing angles

One of my students showed me a problem that she says is similar to what they would do in high school in her home country (which i am attaching here. . The goal of the problem is to find the measure of $$\angle DEC$$ using the given angle measures provided. I've tried working on this to see what other angle measures I could deduce, and I'm including that here This is where I'm stuck. I've tried:

• Labeled one unknown angle as x and determined all other unknown angles in terms of $$x$$, but it's fully consistent and nothing seems to simplify to indicate what $$x$$ is.
• Drawn in lines parallel to the sides through various points, and use what I know about parallel lines cut by transversals, but it doesn't seem to get me any closer to the target.

I suspect I might need to draw in some additional line or extend the diagram in some way but I can't figure out what. Any help would be greatly appreciated. Thanks!

• +1 Interesting problem, nicely presented, good work shown. One constructive criticism: Please replace the graphic labeled "Image of Problem" to a similar graphic written in English. mathSE reviewers (like me) will want to try to absorb the original question; this absorption is impeded re the language. Oct 9 '20 at 22:22
• the way to go may be with sine law Oct 9 '20 at 23:07
• Maths in Chinese senior high schools (高中) is notoriously difficult. Oct 10 '20 at 3:55

Extend $$BC$$ to $$F$$ such that $$∠BFA = ∠DEB$$.

Then, $$AEDF$$ is a cyclic quadrilateral.

$$∠DAF = 180° - 30° - 60° = 90°$$

Therefore, the center of the circle AEDF should lie on DF. Let O be the center.

$$∠AOC = 2∠AFD = 60° = ∠ODG$$

$$AO = OD$$

$$∠OAC = ∠OAD - ∠CAD = 20° = 2∠DAE = ∠DOG$$

Then, $$△OAC≅△DOG$$.

$$∠DEG = 180° - 20° - ∠DGE = 160° - 60° -20° = 80° = ∠DGE$$

Hence, $$OC = DE$$. Moreover, $$∠EOC = 2(10°) = ∠ADE$$ and $$OE = DA$$.

Then, $$△EOC≅△ADE$$.

$$∠DEC = 180° - 30° - ∠CEO - ∠OEA = 150° - ∠EAD - 70° = 70°$$

I used this as a last resort, and I am sure that a more elegant solution exists, but this is by coordinate geometry. Set the base of ABC as the x axis, and the altitude from said base to be the y axis. The lines AC and CB then have equations: $$\tan\left(50\right)x+\sin50=y$$ $$-\tan\left(50\right)x+\sin50=y$$ Since liens AB and AD share a x intercept of $$\cos50$$ and AD has a slope of $$\tan(10)$$, we can derive the equation of AD to be: $$\tan\left(10\right)\left(x+\cos50\right)=y$$ Lines AD, ED and CB intersect at D, and ED has a slope of $$\tan(10+20)=\tan(30)$$, from there we can deduce that the equation for ED by finding the coordinates to point D and using the point slope form of a line: $$\tan(10)x+\tan(10)\cos(50)=-\tan(50)x+sin(50)$$ $$x_D=\frac{\left(\sin50-\tan\left(10\right)\cos\left(50\right)\right)}{\tan\left(10\right)+\tan\left(50\right)}$$ $$y_D=-x_D\tan(50)+\sin(50)$$ The equation of ED: $$\tan\left(30\right)\left(x-x_{d}\right)-x_{d}\tan\left(50\right)+\sin50=y$$ The line CE shares a x intercept with ED at E and also shares a y intercept with AC and CB at C. From these two points we can figure out the slope of ED. Notice that we only need the slope, because the aim is to find an angle. We do not need the whole equation.

$$x_E=\frac{x_{d}\tan\left(50\right)-\sin\left(50\right)}{\tan30}+x_{d}$$ $$y_E=0$$ $$x_C=0$$ $$y_C=\sin50$$ $$slope=\frac{\Delta x}{\Delta y} = \frac{-\sin50}{\frac{x_{d}\tan\left(50\right)-\sin\left(50\right)}{\tan30}+x_{d}}$$ Therefore: $$\arctan(\frac{-\sin50}{\frac{x_{d}\tan\left(50\right)-\sin\left(50\right)}{\tan30}+x_{d}})$$ is the angle that CE makes with the horizontal, this turns out to be exactly -80 degrees. If we put that back into the context of the problem, that means that $$\angle BEC$$ is $$180-80=100$$ degrees, and since $$\angle DEC= \angle BEC - \angle BED$$ and $$\angle BED$$ is 30 degrees. we can conclude $$\angle BEC = 100 - 30 = 70^\circ$$

Note, I omitted a lot of the ugly algebra, for example, it is not rigorous to conclude that the slope of CE is 80 degrees simply with a calculator, it is important that you use trigonometric identities to verify for yourself.

Here's the proof with sine rule. This will be the shortest solution, but it is inelegant, as is the case with all "World's Hardest Easy Geometry Problem" type problems. And as with all those questions, there will be a nice solution, (hopefully.) Consider the "simplified" diagram below.

I messed up the naming so we are looking for $$\theta = \angle EDC$$.

In $$\triangle CDE: \dfrac {EC}{\sin \theta} = \dfrac {DC}{\sin (180^\circ - 80^\circ - \theta)} = \dfrac {DC}{\sin (80^\circ + \theta)}$$

In $$\triangle ACE: \dfrac {EC}{\sin 40^\circ} = \dfrac {AC}{\sin 80^\circ}$$

In $$\triangle ACD: \dfrac {AC}{\sin 150^\circ} = \dfrac {DC}{\sin 10^\circ}$$

Hence:

$$\frac {\sin (80^\circ + \theta)}{\sin \theta} = \frac {DC}{EC} = \frac {AC \sin 10^\circ}{\sin 150^\circ} \cdot \frac {\sin 80^\circ}{AC \sin 40^\circ} = \frac {\sin 10^\circ \sin 80^\circ}{\sin 150^\circ \sin 40^\circ}$$

$$\frac{\sin (80^\circ + \theta)}{\sin \theta} = \frac {\sin 80^\circ \cos \theta + \cos 80^\circ \sin \theta}{\sin \theta} = \sin 80^\circ \cot \theta + \cos 80^\circ$$

Thus we have:

\begin{align}\theta &= \cot^{-1} \left(\frac {\sin 10^\circ}{\sin 150^\circ \sin 40^\circ}-\cot 80^\circ\right)\\ & = \cot^{-1} \left(\frac {\sin 10^\circ}{\sin 30^\circ \sin 40^\circ}-\tan 10^\circ\right)\\ & = \cot^{-1} \left(\frac {2\sin 10^\circ}{4\sin10^\circ \cos 10^\circ \cos 20^\circ}-\frac{\sin 10^\circ}{\cos 10^\circ}\right)\\ & = \cot^{-1} \left(\frac {1}{2\cos 10^\circ \cos 20^\circ}-\frac{\sin 10^\circ \cos 20^\circ}{\cos 10^\circ \cos 20^\circ}\right)\\ & = \cot^{-1} \left(\frac {\sin 30^\circ - \sin 10^\circ \cos 20^\circ}{\cos 10^\circ \cos 20^\circ}\right)\\ & = \cot^{-1} \left(\frac {\cos 10^\circ \sin 20^\circ}{\cos 10^\circ \cos 20^\circ}\right)\\ & = \cot^{-1} \tan 20^\circ\\ & = 70^\circ \end{align}

The last steps are only possible if we know in advance the solution is nice.

• may I ask you what software did you use to produce the image, I would like have such a software, Oct 11 '20 at 16:20
• @Physor It's GeoGebra Classic 6. It's free. Oct 11 '20 at 17:04
• Thanks a lot for the reply!! Oct 11 '20 at 17:04

The problem looks set up for a couple applications of the m-n cot theorem. Let $$\angle DEC=\theta$$.
By that theorem and using the angles you have already marked in your second image, we have the following two equations $$\text{ In }\triangle ABC\text{ with cevian } AD, \quad (BD+CD)\cot(60^\circ)=BD\cot(10^\circ)-CD\cot(40^\circ) \qquad(1)\\ \text{In }\triangle EBC\text{ with cevian } ED, \quad (BD+CD)\cot(80^\circ)=BD\cot(30^\circ)-CD\cot(\theta) \qquad(2)$$

Take $$\dfrac{BD}{CD}=z$$, divide $$(1),(2)$$ by $$CD$$ on both sides, you get $$\dfrac{z+1}{\sqrt 3}=z\cot(10^\circ)-\cot(30^\circ+10^\circ) \qquad(3)\\ \quad (z+1)\cot(90^\circ-10^\circ)=z\cot(30^\circ)-\cot(\theta) \qquad(4)$$

Now, we can just eliminate $$z$$ from the two equations to find the other unknown quantity $$\theta$$.
Let $$\cot(10^\circ)=t$$.
$$\cot(40^\circ)=\cot(30^\circ+10^\circ)=\dfrac{\cot(30^\circ)\cot(10^\circ)-1}{\cot(10^\circ)+\cot(30^\circ)}=\dfrac{\sqrt3t-1}{\sqrt 3+t}$$ $$\cot(80^\circ)=\cot(90^\circ-10^\circ)=\tan(10^\circ)=\dfrac1t$$

Eliminating $$z$$ easily gives $$\cot(\theta)=\dfrac{\frac{t}{\sqrt3}-1}{t+\frac1{\sqrt3}}=\dfrac{\cot(10^\circ)\cot(60^\circ)-1}{\cot(10^\circ)+\cot(60^\circ)}\\=\cot(60^\circ+10^\circ)=\cot(70^\circ) \implies \theta=70^\circ$$

Let AB=2 units $$\dfrac{AB}{\sin ADB}=\dfrac{2}{\sin 120^0}=\dfrac{4}{\sqrt 3}$$ $$BD= \sin 10^0 *\dfrac{4}{\sqrt 3}$$ $$\dfrac{\sin50^D}{DE}$$ Calculate DE from above $$DC=BC-DB\;$$