Let the vertex of an angle $ABC$ be located outside a circle and let the sides of the angle intersect equal chords $AD$ and $CE$ with the circle. Prove that the angle $ABC$ is equal to the half the difference of the angles subtended by the chords $AC$ and $DE$ at the center.


  • $\begingroup$ Welcome to SE. You'll get a better response if you show some effort on your own part to find a solution. $\endgroup$ – Logophobic Feb 12 '16 at 16:58
  • $\begingroup$ Could you provide a drawing of the statement? $\endgroup$ – Octania Feb 12 '16 at 19:02
  • $\begingroup$ what are you allowed to use to prove this ? This is usually used as a theorem $\endgroup$ – Svetoslav Feb 12 '16 at 19:20
  • $\begingroup$ W.l.o.g. consider the unit circle, with $$A=\begin{pmatrix}\cos\alpha\\\sin\alpha\end{pmatrix}\quad C=\begin{pmatrix}\cos\gamma\\\sin\gamma\end{pmatrix}\quad D=\begin{pmatrix}\cos\delta\\\sin\delta\end{pmatrix}\quad E=\begin{pmatrix}\cos\varepsilon\\\sin\varepsilon\end{pmatrix}$$ and looking at the angle between $$\begin{pmatrix}\sin\alpha-\sin\delta\\ \cos\delta-\cos\alpha\end{pmatrix}\text{ and }\begin{pmatrix}\sin\gamma-\sin\varepsilon\\ \cos\varepsilon-\cos\gamma\end{pmatrix}$$ using trig. identities $\endgroup$ – MvG Feb 12 '16 at 20:57
  • $\begingroup$ … At least that's what I thought, since I do like coordinates. But so far I haven't found the right combinations of identities to get rid of the angles subtended by $AD$ and $CE$. So if someone else wants to give this route a try, I'd wish them the best of luck with this approach. $\endgroup$ – MvG Feb 12 '16 at 21:01

I like homogeneous coordinates and the tangent half-angle substitution. So w.l.o.g. the circle is the unit circle, with $A=(\cos\alpha,\sin\alpha)$ and so on for the other three points. With $a:=\tan\frac\alpha2$ you get $A=[1-a^2:2a:1+a^2]$ in homogeneous coordinates. Likewise for $c,d,e$ defining $C,D,E$. The line joining two points can be computed using the cross product of these homogeneous coordinate vectors, and the first two coordinates of that are the normal vector of the line. So the line $AD$ has normal vector $\bigl((1-ad)(a-d),a^2-d^2\bigr)$ and likewise $CE$ has $\bigl((1-ce)(c-e),c^2-e^2\bigr)$. The angle between the lines at $B$ is the angle between these two normal vectors.

Now as I explained here, the dot product between two vectors is proportional to the cosine of the angle, while the determinant is proportional to the sine. The factor of proportionality is the same in both cases, namely the product of the lengths of the vectors. The tangens of an angle is therefore the determinant divided by the dot product. So you get

\begin{align*} \tan\angle ABC&=\frac{ \begin{pmatrix}(1-ad)(a-d)\\a^2-d^2\end{pmatrix}\cdot \begin{pmatrix}(1-ce)(c-e)\\c^2-e^2\end{pmatrix}}{\begin{vmatrix} (1-ad)(a-d) & (1-ce)(c-e) \\ a^2-d^2 & c^2-e^2 \end{vmatrix}} \\[3ex] &=\frac{- a c d + a c e - a d e + c d e - a + c - d + e} {a c d e + a c - a d + c d + a e - c e + d e + 1} \end{align*}

That's what my computer algebra system says. Now for the other half of the task. There you are talking about differences between angles, so recall


Let's define $$s(x,y):=\frac{x-y}{1+xy}$$ as a function where you can compute the difference of angles while actually operating on their tangens. Remember that $a,c,d,e$ were the tangens of half the polar angle of these four points, so $s(c,a)$ is the tangens of half the angle subtended by $AC$. Likewise $s(d,e)$, and the tangens of the difference between these two angles is $s(s(c,a),s(d,e))$. Some more computer algebra verifies that this is exactly the quotient given above. q.e.d.


Construct the line through $E$ parallel to $AD$, intersecting the circle at $G$:

enter image description here

I assume you already have the theorem about the arc subtended by an inscribed angle, so $$ \angle CEG = \frac12 \angle CFG. $$ The triangles $\triangle AFD$ and $\triangle EFG$ are isoceles, so $\angle ADF = \angle DAF$ and $\angle GEF = \angle EGF$. Can you use those facts to show that $ \angle DFE = \angle AFG? $

Can you see now how to show that $$ \angle ABC = \frac12 (\angle AFC - \angle DFE), $$ which is what you need to prove?


$$\angle ABC = \pi - \angle ACE - \angle DAC = \frac{1}{2}(2\pi - \overset{\frown}{ADE} - \overset{\frown}{DEC}) = \frac{1}{2}(\overset{\frown}{AC} - \overset{\frown}{DE}).$$


Triangles $FDA$ and $FCE$ are mutual images by the orthogonal symmetry with respect to the internal line bissector $(L)$ of angle $(EF,ED)$. As straight lines $ED$ and $EF$ make the same angle wrt to $(L)$, their intersection point, $B$, is on $(L)$.

This symmetry with respect to $BF$ allows to restrict our reasoning to triangle $BCG$ (where $G$ is the intersection of line $BF$ with the circle which is the farthest from $B$).

The following proof is "à la Euclid", by ''angle chasing''.

Let us give names to the following (unsigned) angles:

a = angle FBE, b = angle GFC, c = angle EFB,

f = angle FEC = angle ECF (because triangle FCE is isosceles)

g = angle FGC = angle FCG (because triangle FGC is isosceles as well).

Thus, we have the following relationships:

In triangle $EFC$: $ \ 2f+(\pi-(b+c))=\pi \ $. Thus $ \ f=\frac{b+c}{2} \ \ (1)$.

In triangle $FGC$: $ \ 2g=\pi-b \ \ \ (2)$.

In triangle $BGC$: $ \ a+f+2g=\pi \ \ \ (3)$

Substracting (3) - (2): $ \ a+f=b \ \ \ (4)$

Taking (1) into account, one gets

$$a = \frac{b-c}{2}$$

which is equivalent to the desired relationship.

Tthe different (unsigned) angles.


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