I'm confused at which property or identity can be used to find the angle in a triangle when it looks inscribed in a circle but one of its sides doesn't appear to pass through the center. I'm also confused about the other angles, particularly to the ones which are near the tangent line.

The problem is as follows:

A certain serum protein is under research for its optical properties in a medical laboratory in Taichung. The results shown a circumference whose points $A$, $B$, $C$, $D$ and $E$ represent the atoms of the protein crystal. In order to find the protein's optical properties a technician uses a set of beams which draw lines and form angles in the circumference. One beam is tangent to the circumference, while the others drawn an angle labeled as $\omega$ which are congruent. What would be the angle represented by $\phi$?

Sketch of the problem

The given alternatives in my book are:

$\begin{array}{ll} 1.& 36^{\circ} \\ 2.& 45^{\circ} \\ 3.& 53^{\circ} \\ 4.& 72^{\circ} \\ 5.& 108^{\circ} \\ \end{array}$

What I tried to do is to draw the known information in the sketch as shown below.

Sketch of the solution

Since the line (beam) is tangent to the circumference, then this should make a $180^{\circ}$ angle with it. So I assumed all the $\omega$ angles cover the whole tangent beam colored with blueberry therefore:




Sketch of the solution

Then I traced a chord between $CD$ and from then the other identity which I could identify was that the arc given between $\overset{\frown}{AED}$ covers the angle on $\angle ABD$ and $\angle ACD$ therefore:

$\angle ABD = \angle ACD$

Now the part where I'm still doubtful is if there was a way to prove that $\angle ADB = \angle BAD = \angle CDB$ ?

The only thing that I could come up with was to establish that the segment $AB = CD$. But this is which I don't know if its right. By doing this triangle congruence can be used and state that the angles mentioned earlier are the same, but again I don't know if that's right.

Assuming that what I did was okay, the rest would be just using the triangle formula for the sum of its interior angles as follows:

$2\omega + \phi + \omega = 180^{\circ}$



Which correspond to the fourth alternative. But as mentioned I'm not very sure if what I did was the right thing. Therefore. Can somebody help me with this and clear out my doubts?.


Well, as you know, $\omega=36^\circ$. Now, the bottom two $\omega$’s and $\phi$ cut out the same arc of the circle, namely the arc $AED$, so that $\phi=2\omega=72^\circ$.

  • $\begingroup$ This is probably the simplest way to get the answer. Worth noting that the justification of $\phi=2\omega$ is the “inscribed angle theorem.” en.wikipedia.org/wiki/Inscribed_angle $\endgroup$
    – Steve Kass
    Mar 21 '19 at 2:13
  • $\begingroup$ Well, @SteveKass, I learned it in high school about 70 years ago. Don’t they still teach it? $\endgroup$
    – Lubin
    Mar 21 '19 at 3:49
  • $\begingroup$ Well, they did nearly 50 years ago, when I was in high school. I thought a reference might be helpful for readers who didn’t know that a key detail was that the vertex of each angle was on the circle. $\endgroup$
    – Steve Kass
    Mar 21 '19 at 15:11

Good try! The solution you get is actually correct, but I think the path you took is not flawed.

I agree with your calculation of $\omega = 36^\circ$ and indeed $\angle ABD = \angle ACD=\varphi$.

Now the answer to your question is no, $\angle ADB $ and $\angle BAD$ are different, but I think you know that yourself. You now say that you can establish $AB = CD$. How do you justify this? Later, this turns out to be correct, but I can not find a justification for that just now.

Instead you can use the symmetry of the situation to see that $\overline{AC}=\overline{AD}$, (for a rigorous proof of this fact, see Lemma at the end.) Now the triangle $ADC$ is an isosceles triangle and thus $\angle ACD = \angle ADC$.

enter image description here

Now you can see that $$ \omega + 2 \varphi = 180^{\circ}, $$ and thus indeed $\varphi = 72^{\circ}$.

Lemma: Let $A, C, D$ be three points on a circle with $\angle MAC = \angle MAD$. Then $\overline{AC} = \overline{AD}$.

Proof Consider the following image.

enter image description here

The triangles $AMC$ and $AMD$ are isosceles, since two of their sides have length $r$. Thus the green angles are are congruent to the blue angles, i.e. $\angle MAC = \angle MAD = \angle MCA = \angle MDA$. Since two angles in the triangles $AMC$, $AMD$ are congruent, all three are congruent, and thus the triangles are congruent and $\overline{AC} = \overline{AD}$.

  • $\begingroup$ Sorry, I didn't typed correctly my question. What I meant was if there was a way to prove if $\angle BAD = \angle ADC$ or in other words if $\angle CAD = \angle BAD$ ? because I've tried every trick under my sleeve and I can't find a way to prove it. Its true, I'm still doubtful on how to justify $AB=CD$. I just taken for granted. The thing with symmetry is where I'm still confused. How can I prove that the segment you traced from A dividing the circle does produce an exact copy of the upper triangle so that it makes $\angle ADC = \angle ACD$?. $\endgroup$ Mar 19 '19 at 22:57
  • $\begingroup$ I mean can I use that property of symmetry so freely? I hope you can help me with these questions, please!. $\endgroup$ Mar 19 '19 at 22:58
  • $\begingroup$ Dear Chris, I think there might be a typo in your question again. I do not know how to prove $\angle BAD = \angle ADC$ directly. As far as I can see this follows from the fact that $\angle ACD = \angle ADC$. Similarly, I do not see how to prove $\overline{AB}=\overline{CD}$ directly. I do feel very comfortable using the symmetry 'freely'. But I have added a proof in the form of a lemma for you. Let me know if you need more clarification. $\endgroup$ Mar 21 '19 at 0:31

Inscribe a pentagon in a circle with points A, B, C, D, E.
Place a point O at the center.
Calculate the central angles AOB, AOC, BOC.
From there calculate angles ABC, BCA, CAB.
Do likewise for the other sides.
Compare the result with the diagram you have presented.

  • $\begingroup$ I tried to follow your approach but I don't know how to calculate the central angles you mentioned. Although the situation would be eased if the resulting pentagon would be regular but in this case it doesn't seem this way. Can you perhaps add some drawing or another hint to help me calculate the angles you mentioned?. The only thing I could come up with was $\angle BOC = \angle BAC$. I'm assuming $AD$ does pass through the center. Can you expand your help please? $\endgroup$ Mar 19 '19 at 23:09
  • $\begingroup$ The central angle AOB is 360/5 degrees. The angles ABO and BAO are equal as AOB is an isoscoles triangle. All three angles of AOB sum to 180 degrees. After you have determined all the angles including the angles BAP, EAQ where P and Q are points on the tangent at A, you can determine if your diagram and the inscribed five pointed are similar. $\endgroup$ Mar 20 '19 at 7:26

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