$\triangle ABC$ has circumcenter $O$; $BO$ and $CO$ meet $AC$ and $AB$ at $D$ and $E$. If $\angle A=\angle EDA=\angle BDE$, show they are $50^\circ$

There is surely a purely geometrical solution to this problem, but none forthcoming so far! Trigonometry confirms the result, however the quest is to solve this geometrically, almost certainly involving a clever construction, leading to an equilateral triangle and hence exposing angle values. So a possible nod to “Adventitious Angles”….

… OR, could it be possible to prove the impossibility of a purely geometrical proof?

$$O$$ is the circumcenter of $$\triangle ABC$$. $$BO$$ cuts $$AC$$ at $$D$$, and $$CO$$ cuts $$AB$$ at $$E$$, as in the figure. Angles $$EAD$$, $$EDA$$, and $$BDE$$ are equal. Prove that their value is $$50^\circ$$

You can quickly work out all the angles in terms of x and y=angle OBC and discover that x+y=90. Also that triangle COD is isosceles and further that it’s sufficient to prove triangle BCE isosceles to get the answer.

This may help ... repeat “may”! Define point F on AD such that CE = CF. Then since triangle COD is isosceles, then (among other things) EF is parallel to OD. Angle BCF will ultimately be shown to be 60 degrees, so we need to show that triangle BFC is equilateral. Worth exploring a bit; I’m convinced the key is the equilateral triangle because this forces another relationship between x and y.

UPDATE:- I have made a bit of progress via the construction of the line BF, as described above. Since triangle COD is isosceles and CF=CE by construction, then triangle CEF is also isosceles and is similar to triangle COD. The following can then be proved:-

1. Angle EFD = 180-2x
2. Angle FED = x
3. FE=FD
4. FD=EO
5. triangle EFA is congruent to triangle EDO
6. triangle EFA is similar to triangle ABD

Now construct the line FO and drop a perpendicular from O to point G on BC. Then the following can be proved:

1. FO bisects angle EOD into x + x and also bisects angle FB into (180-3x) + (180- 3x)
2. The points F O and G are colinear
3. BG = GC
4. Triangles CFG and GFB are congruent
5. CF = FB
6. Triangle CFB is isosceles with CF = FB

BUT .... we still need to prove that triangle CFB is equilateral, which is very elusive. There are two distinct ideas here… SO ... any help is greatly appreciated! Thanks for your interest.

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– Blue
Commented Jun 3, 2022 at 12:08
• Triangle $COD$ is not an isosceles triangle. But triangle $COB$ is an isosceles triangle.
– YNK
Commented Jun 4, 2022 at 9:52
• FOR YNK:- Sorry, but they are both isosceles triangles. COB is (as you say) because two sides are radii of the circle. However COD is because:- 1. Angle COB=2x (angle at centre of circle by chord CB) 2. Therefore angle COD=180-2x. 3. Angle ADB=2x. 4. Therefore angle ODC =180-2x. 5. So triangle COD is isosceles because of two equal angles. Commented Jun 4, 2022 at 10:07
• Yes, you are right. I made a mistake. So, I apologize.
– YNK
Commented Jun 4, 2022 at 15:25
• @wotsit: Another tip: To address someone in a comment, put @ before their name (as I just did with this one). This will send them a notification (as you just received) to get their attention. ... This only works with one person per comment, and it only works if that person has participated in comments.
– Blue
Commented Jun 4, 2022 at 18:48

Let $$AO$$ cut $$BC$$ at $$F$$. It is well-known that the quadruple $$(DA, DB; DE, DF)$$ is harmonic. Since $$DE$$ is the internal angle bisector of the angle $$ADB$$, it follows that $$DF$$ is the external angle bisector of $$ADB$$. In particular $$\angle ODF = \frac\pi2 -\frac12\angle ADB = \frac\pi2-\angle A=\angle OCF.$$ It follows that the quadrilateral $$CFOD$$ is cyclic. From this we get $$\pi=\angle C +\angle FOD =\angle C + 2\angle C=3\angle C,$$ hence $$\angle C=\frac\pi3$$.

On the other hand we have $$\pi-3\angle A=\angle DBA=\frac\pi2-\angle C=\frac\pi2-\frac\pi3=\frac\pi6.$$ Hence $$\angle A=\frac13\left(\pi-\frac\pi6\right)=\frac{5\pi}{18}.$$

• This is a very elegant solution, for which I am most grateful. However, please forgive me if I ask a few clarification questions, particularly if they appear obvious. 1. Where do you use the fact that quadruple (𝐷𝐴,𝐷𝐵;𝐷𝐸,𝐷𝐹)is harmonic? 2. Why is 𝐷𝐹 the external angle bisector of 𝐴𝐷𝐵? 3.Why does angle OCF equal 90 - angle A? 4. Why does angle FOC equal 2 x angle C. 5. Why does angle DBA equal 90 - angle C. Grateful for explanation.. Many thanks. Commented Jul 25, 2022 at 16:55
• Q4 should read “why does angle FOD equal 2 x angle C?” Apologies. Commented Jul 25, 2022 at 17:32
• @wotsit 1 and 2: it is a standard lemma that "harmonic"+"internal bisector" implies "external bisector". You can prove it with Ceva, Menelaos, and angle bisector theorem. Commented Jul 25, 2022 at 21:08
• @wotsit 3,4,5: use the fact that the central angle equals twice the inscribed angle, and that in an isosceles triangle the angles at the base equal 90 degrees minus half of the angle at the vertex opposite the base. Commented Jul 25, 2022 at 21:11
• Perfect; thanks. 3,4,5 were obvious, but 1,2 form the key to this puzzle. Much appreciate your work! Commented Jul 25, 2022 at 22:03

The following solutions goes as much as possible geometrically. However, at some point the trigonometry has to enter the game, else i cannot finish. (After having the trigonometric equation and solution one may extract from it why it is not obvious to have a synthetic solution. The trigonometric equations are transformed very detailed, this is so since after some tired eyes error i was parallely checking them with a computer.) For problems like this one, where a lot of points and properties of them are given "in the same time" i am always trying to put order in the structure, and find a way where points come with associated properties one by one. Usually, this leads to a solution. Here also the case. So let's start.

Let us collect some constraints, as they may be extracted from the figure. We try to proceed in a constructive manner, introducing points one by one in terms of the only parameter $$x$$. We cannot start with the points $$A,B,C$$ since the condition of getting three $$x$$ angles as in the problem cannot be simply arranged, but we can start with $$A$$, $$D$$ and $$x$$ and construct one by one points from the given figure. Below we have thus two worlds. One of the given figure, "(F)", we isolate properties from it. And one world "(K)" in which we construct this situation.

We start thus in (K) with the points $$A$$ and $$D$$ and with the angle $$x$$. Then the point $$E$$ is uniquely determined to match the given figure. Let us fix ideas and norm $$EA=ED=1$$. The point $$B$$ is also determined as intersection of the ray $$AE$$ with the ray in $$D$$ obtained by reflecting $$DA$$ w.r.t. $$DE$$, thus insuring the two $$x$$-angles in $$D$$.

In (F) the angle $$\widehat{ABD}$$ from $$\Delta ABD$$ is equal to $$180^\circ -3x$$. In particular, $$x$$ is between $$0^\circ$$ and $$60^\circ$$ so far. The angle $$\widehat{BOC}$$ is $$2\cdot \widehat{BAC}=2x$$. This gives in $$\Delta DEO$$ the angle $$\widehat{DEO}=180^\circ - 3x=\widehat{BOC}$$. In particular the two triangles are similar: $$\Delta DEB$$ and $$\Delta DOE$$. So the circle $$\odot(BEO)$$ is tangent in $$E$$ to $$ED$$. Let us denote by $$\Gamma$$ the center of this circle. This property can be used to construct $$O$$.

In (K) we consider the intersection of the perpendicular in $$E$$ on $$ED$$ with the perpendicular bisector of the segment $$EB$$. Denote it by $$\Gamma$$. Draw the circle $$\odot(\Gamma)$$ centered in $$\Gamma$$ with radius $$\Gamma B=\Gamma E$$. It intersects $$DB$$ for the second time in a point $$O\ne B$$. We finally construct $$C=AD\cap EO$$. It's time for a picture of (K):

The marked angles are determined one by one, $$\widehat {DEB}=2x$$, then $$\widehat {\Gamma EB}=2x-90^\circ$$ (with sign convention), then $$\widehat {E\Gamma O}=2\widehat {EBO}=2(180^\circ-3x)$$, then $$\widehat {\Gamma OE}=3x-90^\circ$$ in $$\Delta \Gamma OE$$.

The angle $$\widehat {\Gamma OB}=\widehat {\Gamma BO}$$ is $$270^\circ -5x$$, the angle that makes the sum $$2(2x-90^\circ)+2(3x-90^\circ)+2(270^\circ-5x)$$ equal to $$180^\circ$$, the sum of the angles in $$\Delta BEO$$. Or the angle that makes $$\widehat{BOD}=180^\circ$$, given the two angles $$\widehat {EOD}=2x$$, $$\widehat {EO\Gamma}=3x-90^\circ$$.

And in this situation we should have $$OA=OB$$. (The we can check the full relation $$OA=OB=OC$$, since $$O$$ would be on the side bisector of $$AB$$, and on the arc through $$BC$$ having the geometric locus property $$\widehat{BOC}=2x$$. So we can discard $$C$$ from the discussion.)

From the condition $$OA=OB$$ we search a constraint for $$x$$. The condition of having $$O$$ "as in the figure" in the interior of $$\Delta ABC$$ leads to $$180^\circ-3x= \widehat{OBA} = \widehat{OAB} < \widehat{DAB} = x\ ,$$ so $$180^\circ<4x$$, giving a lower bound $$45^\circ$$ for $$x$$. This determines $$\widehat{OAD} = x - (180^\circ-3x)= 4x - 180^\circ$$. To see the level of complication put on the road in the problem, let us go in the trigonometry with $$x$$. The sine theorem and the similarity mentioned above give the following relations - recall $$EA=ED=1$$, $$AD=2\cos x$$: \begin{aligned} &\Delta DEB\sim\Delta DOE && 1^2 = DE^2 = DO\cdot DB\ , \\ &\Delta EOD && \frac{\sin 2x}1 = \frac{\sin x}{OE} = \frac{\sin 3x}{OD} \ , \\ \\ &\Delta AOD && \frac{\sin 2x}{AO} = \frac{\sin (4x-180^\circ)}{OD} \overset!= \frac{\sin (360^\circ-6x)}{AD} \ . \end{aligned} From here, the two bang (exclamation mark) equalities, the one above together with $$OB \overset!= OA$$, lead to two (possibly equivalent - but i am happy to have and use both) equations: \begin{aligned} &\frac{\quad-\sin 4x\quad}{\displaystyle\frac{\sin 3x}{\sin 2x}} = \frac{-\sin 6x}{2\cos x}\qquad\text{ and} \\[2mm] & \frac{\sin 2x}{\sin 3x} - \frac{\sin 3x}{\sin 2x} = DB- DO =OB \overset!= OA = -\frac{\sin 2x\cdot 2\cos x}{\sin 6x} = -\frac{\sin 2x\cos x}{\sin 3x\cos 3x} \ . \end{aligned} In the last equation there is a simplification with $$\sin 3x$$ in denominator. We get successively: \begin{aligned} &\cos 3x\cdot 2\Big(\sin^2 2x - \sin^2 3x\Big) = - 2\sin^2 2x\cos x\ , \\ &\cos 3x\cdot\Big( (2\cos^2 3x -1)- (2\cos^2 2x-1)\Big) = - 2\sin^2 2x\cos x\ , \\ &\cos 3x\cdot\Big( \cos 6x- \cos 4x\Big) = - 2\sin^2 2x\cos x\ , \\ &-\cos 3x\cdot2\sin 5x\sin x = - 2\sin 2x\cdot \sin x\cdot 2\cos^2 x\ , \\ &2\cos 3x\sin 5x = 2\sin 2x\cdot (\cos 2x+1) \ , \\ &\sin 8x + \sin 2x= \sin 4x + 2\sin 2x \ , \\ &\color{blue}{\sin 8x = \sin 4x + \sin 2x} \ . \end{aligned} The other equation gives after a fist step of multiplication with $$\sin x$$: \begin{aligned} &2\sin 4x\sin 2x\cdot 2\cos x = 2\sin 3x\sin 6x\ , \\ &2\sin 4x\sin 2x\cdot \sin 2x = 2\sin x\sin 3x\sin 6x\ , \\ &\sin 4x(1-\cos 4x) = (\cos 2x-\cos 4x)\sin 6x\ , \\ &2\sin 4x -\sin 8x = (\sin 8x+\sin 4x) - (\sin 10x+\sin 2x)\ , \\ &2\sin 8x = \sin 2x+\sin 4x+\sin 10x\ . \\ \end{aligned} Combining the two equations we get $$\sin 8x=\sin 10x$$, then $$2\cos 9x\sin x=0$$, so $$\color{blue}{\cos 9x=0}\ .$$ The angle $$9x$$ is between $$9\cdot 45^\circ=405^\circ$$ and $$9\cdot 60^\circ=540^\circ$$, and the cosine vanishes only in $$450^\circ$$ in between, so $$9x=450^\circ$$, so $$x=50^\circ$$. (This is a necessary condition. It may still lead to a contradiction, but we have shown the direction from the question. More on the validity is postponed.)

$$\square$$

The problem is solved, but some notes may be relevant in case a "purely geometric" solution is wanted.

Note: In the above picture we have an alternative triangle, $$\Delta \Gamma EO$$ which is equilateral for the $$x$$-value in the solution. Showing this is however not simple.

Note: If we start with $$x=50^\circ$$ in the above construction, then it is relatively easy to show that $$OA=OB$$, and that after constructing $$C$$ as $$EO\cap AD$$ we have the full chain $$OA=OB=OC$$.

Here is the construction of points one by one, so that the above result becomes transparent. Fix $$A,D$$ and draw rays building angles of measure $$x=50^\circ$$ on $$AD$$ in $$A$$ and $$D$$. They intersect in a point $$E$$, so that $$\Delta EAD$$ is isosceles in $$E$$. Construct $$B$$ by intersecting $$AE$$ with the reflection of $$DA$$ in $$DE$$. Construct $$O,O'$$ in this order on $$DB$$ with reflections $$\Omega,\Omega'$$ in $$DA$$ w.r.t. the symmetry line $$DE$$ so that $$\Delta EOO'$$ and $$\Delta E\Omega\Omega'$$ are $$80^\circ-80^\circ-20^\circ$$ isosceles triangles in $$E$$, so $$EO'=EO=E\Omega=E\Omega'$$. Then all the angles around $$E$$ can be computed. Construct also $$\Gamma$$ as the circumcenter of $$\Delta EOB$$. The angles in $$E$$, respectively $$\Gamma$$ in the isosceles triangles $$EO\Omega$$ and $$EO\Gamma$$ are each of $$60^\circ$$, so these triangles are equilateral. Note that the triangles $$\Delta O'ED$$ and $$\Delta \Omega'ED$$ are isosceles, making $$O'E\Omega'D$$ a rhombus. The points $$\Omega,\Omega'$$ are placed symmetrically w.r.t. the mid point of $$AD$$, so finally $$A\Omega=\Omega'D$$ is also congruent to the whole chain of marked segments. We observe $$\Delta \Omega AO\equiv \Delta \Gamma BO\ ,$$ both isosceles, congruent marked four segments, with an angle of $$140^\circ$$ in $$\Omega$$, respectively $$\Delta$$. (The angle in $$\Omega$$ is known, since we know the angles in $$\Delta \Omega EO$$ and $$\Delta \Omega OD$$.) This gives $$OA=OB$$. Constructing $$C$$ as $$AD\cap EO$$ makes the angle in $$C$$ in $$\Delta OAC$$ a $$20^\circ$$-angle, same as in $$A$$, so $$OA=OB=OC$$.

Note: The two blue equations also have as solution $$x=10^\circ$$ and $$x=70^\circ$$. But they do not lead to solutions for the given geometrical situation.

Not an answer but something which might prove useful in one.

This looks like an adventitious-angles problem, but with the four initial constraints given as two angle-equalities ($$\angle EDA=\angle ODE=\angle CAB$$) and two length-equalities ($$OA=OB=OC$$) rather than explicit angles.

Draw diameter $$AOG$$. Then $$OG=OB$$. It can be found from angle-chasing that $$\angle COG=\angle EOA=8x-360^\circ$$, $$\angle GOB=\angle AOD=360^\circ-6x$$ and $$\angle OBG=\angle BGO=3x-90^\circ$$.

Now if it could be proved that $$BG=OB$$, too, then $$x=50^\circ$$ follows immediately.

• I feel sure this is progress, so many thanks. Triangle BOG is therefore prospectively equilateral which must surely be helpful. I identified one in my opening statement above, so it all chimes with “Adventitious Angles”. Hope this helps. Commented Jun 19, 2022 at 16:37