Looking for alternative proofs of this statement about angles

This is the theorem to prove.

Below is my proof that I consider rather long and complex.

The given data is on this drawing:

• Construct $$\angle DCE = \angle DCB$$. The point $$E$$ on ray $$CE$$ is chosen in such way that $$CE = CB$$, and that is always possible by segment construction axiom.

• Connect points $$B$$, $$E$$, and $$D$$.

• $$_\Delta CDB \cong _\Delta CDE$$ by SAS, because $$CD$$ is their common side, $$CB = CE$$ by construction, $$\angle DCB = \angle DCE$$ by construction. Therefore $$\angle DBC = \angle DEC$$, as they are opposite to side $$CD$$ of these triangles.

• Let $$O$$ be the intersection point of $$AD$$ and $$CE$$. Also connect points $$A$$ and $$C$$ to construct a line $$AC$$.

• $$_\Delta ABC$$ is isosceles because $$AB = BC$$ as given by the statement of the theorem. Then $$\angle BCA = \angle BCA$$.
• Since $$\angle BAD = \angle BCE$$, then $$\angle ECA = \angle DAC$$ as well.

• We have $$AO = CO$$ because $$\angle OAC = \angle OCA$$ implies $$_\Delta AOC$$ is isosceles. As $$AD = CE$$, we get $$AO + OD = CO + OE$$. As $$AO = CO$$, we get $$AO + OD = AO + OE$$, which implies $$OD = OE$$, therefore $$_\Delta OED$$ is also isosceles. Even more, $$\angle OED = \angle ODE = \angle OAC = \angle OCA$$.
• $$_\Delta AED \cong _\Delta CDE$$ by SAS because $$AD = CE$$, $$ED$$ is their common side, and $$\angle ADE = \angle CED$$. This implies $$\angle EAD = \angle ECD$$, therefore $$\angle EAB = 2\alpha - \alpha = \alpha$$, which implies $$AE$$ is on the angle bisector of $$\angle DAB$$.

• $$_\Delta DAB$$ is isosceles since $$AB = AD$$ by theorem's statement. Therefore, angle bisector is also median and altitude. Then median and altitude of $$_\Delta BED$$ to side $$BD$$ also coincide, therefore $$_\Delta BED$$ is isosceles with $$BE = ED$$. Since also $$BD = DE$$, $$_\Delta DEB$$ is equilateral, therefore all internal angles of this triangle are equal to one another, $$60^\circ each$$.

• $$_\Delta AEB \cong _\Delta AED$$ by SAS because $$AE$$ is their common side, $$AB = AD$$, $$\angle EAD = \angle EAB$$. Therefore, $$\angle ABE = \angle ADE$$.

• As points $$B$$ and $$O$$ are both equidistant from points $$A$$ and $$C$$, they lie on perpendicular bisector of $$AC$$. Therefore, $$BO$$ is also median and angle bisector of $$\angle ABC$$. But same goes for $$_\Delta BED$$ and its angle $$\angle EBD$$. Then $$\angle EBO = \angle DBO = 0.5\cdot 60^\circ = 30^\circ$$.

• By using the notation on the last image, $$\angle ABC = \beta + 30^\circ + 30^\circ + \beta = 60^\circ + 2\beta$$.

• Consider the right triangle $$_\Delta BCG$$. Its acute angles sum up to $$90^\circ$$, therefore: \begin{align*} 30^\circ + \beta + \alpha + \alpha + \beta &= 90^\circ \\ 2\alpha + 2\beta &= 60^\circ \\ \alpha + \beta &= 30^\circ \\ \beta &= 30^\circ - \alpha \end{align*}

• Now we are ready to express $$\angle ABC$$:

$$60^\circ + 2\beta = 60^\circ + 2(30^\circ - \alpha) = 60^\circ + 60^\circ - 2\alpha = 120^\circ - 2\alpha$$

Which was to be proven.

What are shorter alternatives to that proof? I'm considering trigonometry, too, but I would prefer good old elementary geometry.

• To check, how pedantic do you want to be / who is your target audience? $\quad$ I would have just stated "symmetry" to substantiate a lot of your steps, which would have shortened the proof significantly (but I get that might not be your intention). IE Connect $AC$, draw the perpendicular bisector $BG$, Reflect $D$ across $BG$ to $E$, observation about the isosceles triangle $BCE$, simple angle chase to get $x = 120^\circ - 2 \alpha$. Apr 21 at 13:31
• @CalvinLin Those who study high school geometry; I'm tidying up my vast collection of geometric facts which I found on the Internet and proved to educate myself. Symmetry can lead to rather advanced concept in geometry, but proofs involving only certain kinds of symmetry (point symmetry and axial symmetry) are welcome! Apr 21 at 13:35
• See the details of how I'd write up the proof that I edited into my comment. Does that satisfy your needs? $\quad$ The "observation about isosceles triangle $BCE$", esp that $CD$ is the perpendicular bisector, if the crux. I believe that argument can be understood by a high school geometry student. Apr 21 at 13:38
• Tried the simplest proof I knew. As a highschool student myself, I guess this can be understood by others
– Gwen
Apr 21 at 15:27

Here's a possible shorter proof. Reflect $$A$$ about $$BD$$ to $$A'$$. As $$\angle DA'B=2\angle DCB$$, then $$A'$$ is the center of the circle through $$BDC$$. Hence $$BA'C$$ is an equilateral triangle and $$DA'C=60°-2\alpha$$.

On the other hand $$DBC={1\over2}DA'C=30°-\alpha$$ and $$ABD=90°-\alpha$$, so that the requested value of $$\angle ABC$$ folows.

• What is the point $E$ in your proof? Apr 21 at 13:50
• Likely a typo. We have $\angle DA'B = 2 \angle DCB$. Apr 21 at 14:05
• @Rusurano A typo, sorry. Corrected now. Apr 21 at 14:29
• $\angle BDC=150^\circ$ from chord $BC$.
– JMP
Apr 21 at 15:58

Trying to present the simplesssst proof. I'll use basic trigonometry and sine rule.

$$AB=BD=a,\angle ABD=2\alpha$$. By simple geometry, trigonometry, $$AD=2a\sin\alpha$$.

Again by sine rule,$$\frac{AD}{\sin\angle ACD}=\frac{AC}{\sin\angle ADC}\implies \sin\angle ADC=\frac{a×\sin\alpha}{2a\sin\alpha}=\frac{1}{2}$$

Thus $$\angle ADC=150°$$(it's obtuse). Then by angle sum property in $$\triangle ABD,\angle BDA=90-\alpha$$.

Finally by angle sum property in quad $$ABDC, x+3\alpha+90-\alpha+150°=360°\implies x=120°-2\alpha$$.

Here's another simple geometrical argument. No trigonometry etc. required.

Draw $$AE$$, of the same length as $$CD$$, bisecting the angle at $$A$$. Now we have ("side-angle-side") congruent triangles $$AEB$$, $$AED$$, $$BCD$$, which means that the triangle $$BDE$$ is equilateral -- all three of its sides are "the third side" of that triangle we have three of.

We can finish the job with a routine angle-chase:

Angle $$x$$ is $$2\beta+60^\circ$$ where $$\beta$$ is the other "small" angle in that repeated triangle. And by looking at the angles around point $$E$$ we see that the "big" angle in the repeated triangle is $$150^\circ$$. So $$\alpha+\beta=30^\circ$$ and so $$x=60^\circ+2\beta=60^\circ+2(30^\circ-\alpha)=120^\circ-2\alpha$$ as required.

(I can't help suspecting there's a slicker way to finish it off -- perhaps involving adding two more copies of that triangle at B and showing that the resulting thing has an angle of $$120^\circ$$ there -- but can't quite see it right now.)

This argument also shows how to construct every diagram of this type: start with an equilateral triangle, draw an external angle bisector of whatever length you like, join the far ends to the other two vertices, and then slap on a third copy of the triangle you've just made two of.

• An image would help Apr 23 at 11:49
• I agree. How's it now? Apr 23 at 13:13
• Much better, +1 Apr 23 at 16:33

Trig approach:

1. Let $$|AB| = a, |AC | = b$$. Let $$\angle ABC = x$$.
2. By angle chasing, $$\angle ACB = \frac{ 180^\circ - x } { 2}$$, $$\angle ACD = \frac{ 180^\circ - x } {2} - \alpha$$, $$\angle ADC = x + 3\alpha$$.
3. By sine rule on $$ABC$$, $$\frac{a}{\sin \frac{ 180^\circ - x } {2} } = \frac{ b} { \sin x }$$, or that $$\frac{a}{b} = \frac{ \cos x/2}{\sin x } = \frac{1}{ 2 \sin x/2}$$
4. By sine rule on $$ADC$$, $$\frac{a}{ \sin \frac{ 180^\circ - x } {2} - \alpha } = \frac{ b}{ \sin x+3 \alpha}$$, or that $$\frac{a}{b} = \frac{ \cos x/2 + \alpha } { \sin x + 3 \alpha }$$.
5. Hence $$\frac{ a}{b} = \frac{1}{ 2 \sin x/2} = \frac{ \cos x/2 + \alpha } { \sin x + 3 \alpha }$$.
6. Show that this is equivalent to $$\sin \alpha ( 2 \cos (x+2\alpha) + 1 ) = 0$$, and hence $$x+2\alpha = 120^\circ$$ as desired.
• $$\Leftrightarrow \sin (x+3\alpha) = 2 \sin x/2 \cos (x/2 + \alpha) = \sin (x + \alpha) - \sin \alpha$$
• $$\Leftrightarrow \sin \alpha = \sin ( x + 3 \alpha) - \sin ( x + \alpha ) = - 2 \sin \alpha \cos ( x + 2\alpha).$$
• $$\Leftrightarrow \sin \alpha ( 2 \cos (x+2\alpha) + 1 ) = 0$$.

Note

• Guessing the form in Step 6 is hard for the high school student. I knew it by factorizing the complex number representation, then sought to trigo it.