If $$AB=AC=EF, AE=BE, \angle BAC=120^{\circ}, \angle ADB=\angle BEF=90^{\circ}$$ find $\angle ABE$

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Following is one method:

Let $\angle ABE=a, \angle AFE=\angle DBE=30^{\circ}-a, \angle EAF=60^{\circ}-a$. Use sine thereom, we have $$\dfrac{\sin{a}}{\sin{(180^{\circ}-2a)}}=\dfrac{AE}{AB}=\dfrac{AE}{EF}=\dfrac{\sin{(30^{\circ}-a)}}{\sin{(60^{\circ}-a)}}$$ $$\Longrightarrow 2\cos{a}\sin{(30^{\circ}-a)}=\sin{(60^{\circ}-a)}$$ $$\Longrightarrow \dfrac{1}{2}+\sin{(30^{\circ}-2a)}=\sin{(60^{\circ}-a)}$$ $$\Longrightarrow 2\cos^2{(a+30^{\circ})}-\cos{(a+30^{\circ})}-\dfrac{1}{2}=0$$ $$\Longrightarrow \cos{(a+30^{\circ})}=\dfrac{\sqrt{5}+1}{4}=\cos{36^{\circ}}$$ so we have $$a=6^{\circ}$$

Question: Can find geometry methods or other simple methods?

  • $\begingroup$ i think you have made a mistake it must be $$\frac{\sin(\alpha)}{\sin(2\alpha)}=\frac{\sin(\pi/6-\alpha)}{\sin(\pi/3-\alpha)}$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 2 '16 at 8:23
  • $\begingroup$ After doing some playing around with Geogebra, it looks like the constraints provided (the two pairs of equal lengths and the given angles) aren't enough to give a unique figure. To get the diagram shown, one must further require that $E$ lie inside $\triangle ABC$. (One arrangement can be found with $E$ somewhere above the triangle, and similarly one can be found with $E$ below this triangle.) So the diagram contains an implicit assumption. $\endgroup$ – Semiclassical Sep 2 '16 at 19:19
  • $\begingroup$ An error in the previous comment: two arrangements can be found when $E$ is below $\triangle ABC$. $\endgroup$ – Semiclassical Sep 2 '16 at 19:30
  • $\begingroup$ Also, $\Delta ABE$ is isosceles, so $\cos a= (\frac{1}{2}AE)/(EF)$. Since $\sin 2a=2\sin a\cos a$, this is equivalent to the equality obtained from the law of sines. So the first use of the law of sines easily can be avoided (though not the second as far as I can see). $\endgroup$ – Semiclassical Sep 2 '16 at 20:02
  • $\begingroup$ @Semiclassical: As you commented, I found that $a=66^\circ$ when $E$ is above the side $AB$ and that $a=6^\circ, 42^\circ$ or $78^\circ$ when $E$ is below the side $AB$. (This comment assumes that $D$ is the midpoint of the side $BC$ and that $F$ is on the line $AD$.) $\endgroup$ – mathlove Sep 11 '16 at 5:03

Let $\angle EBA = \alpha$.

Let $A$ and $X$ be symmetric with respect to $BE$. Then $BX=AB$. Also note that $BE=AE=EX$, so $A,B,X$ lie on the circle with center $E$ and radius $AE$.

Let $Y$ be on $AF$ and $AY=AB$. Since $\angle BAY = 60^\circ$, $\triangle BAY$ is equilateral. Thus $BY=AB$. Therefore $A,X,Y$ lie on the circle with center $B$ and radius $AB$.

Since $AE=BE$ and $AY=BY$, triangles $AYE$, $BYE$ are congruent (sss). In particular $\angle AYE = \angle EYB = \frac 12 \angle AYB = 30^\circ$.

We have $$\angle AXY = \frac 12 (360^\circ - \angle YBA) = 150^\circ = 180^\circ - \angle AYE = \angle EYF.$$

Moreover $$\angle XYA = \frac 12 \angle XBA = \alpha$$ and $$\begin{align*}\angle FEY & = \angle BEY - \angle BEF = \\ & = (180^\circ - \angle YBE - \angle EYB) - 90^\circ = \\ & = 180^\circ - (60^\circ - \alpha) - 30^\circ - 90^\circ = \\ & = \alpha, \end{align*}$$ therefore $$\angle XYA = \angle FEY.$$

It follows that triangles $AXY$ and $FYE$ are similar. Their similarity ratio is $\dfrac{AY}{EF} = 1$ so in fact they are congruent. Thus $EY=XY$.

Using the fact that $EXY$ is isosceles we get \begin{align*}180^\circ & = \angle XYE + 2 \angle YEX = \\ & = 30^\circ + \alpha + 2(\angle YEA - \angle XEA) = \\ & = 30^\circ + \alpha + 2\angle BEY - 2\angle XEA = \\ & = 30^\circ + \alpha + 180^\circ + 2 \alpha - 8 \alpha = \\ & = 210^\circ - 5 \alpha, \end{align*}

therefore $$\alpha=6^\circ.$$


Used symbol $\theta = \angle ABE$ as it is a variable. Agreement with OP's value of $\theta = 6^0$ for $ \alpha = 30^0$ obtained as follows.

al/th MSE prob

$ BEDF$ is a cyclic quadrilateral with $BF$ as diameter. A construction of the circum(semi-)circle is suggested and found helpful.Hand sketch (Paint) is not to scale.Axes are tilted in graph as shown keeping $EF$ horizontal.Triangle $ABE$ is isosceles, the two important angles equal those in alternate semi-circular segment.

A general derivation to relate $ \alpha= \pi/6$ (in this case) to find/relate to $ \theta $ is given :

Coordinates of E: $ (L \tan \theta , L )\tag{1} $

Equation of EG by Polar Normal form for a straight line :

$$ x \cos (\pi/2- \theta) + y \sin (\pi/2- \theta) = L\sec \theta $$

$$ x \sin \theta + y \cos \theta = L\sec \theta \tag{2}$$

Solving (1),(2) by Cramer's Rule two unknowns to get coordinates of $F$.

$$x_F= N_x/\Delta = L(\sec \theta - 2 \cos \theta)/\Delta,\,\, y_F= N_y/\Delta = L(- 2 \sin \theta- \tan \alpha \sec \theta )/\Delta;\,\tag{3} $$


$$\Delta = (\sec \theta - \tan \alpha \cos \theta) $$

$$ EF^2 = (x_F- L \tan \theta )^2 + ( y_F- L)^2 = 4 L^2 \tag{4} $$

$$ (N_x/\Delta-\tan \theta))^2 +(N_y /\Delta-1 )^2 = 4 \tag{**5**} $$

which is the required relation implicit in $\alpha,\theta $. The numerical solution is obtained that could be applied for a range of similar problems. To check out its use that way some solution pairs that have integral degree combinations:

$$ (\alpha, \theta)=( \approx 26.56505,0^0), (30^0,6^0), (36^0,18^0), (40^0,70^0),(45^0,60^0+ 4 \pi)... \tag{6} $$

I was curious about such $ (\pi/n, \pi/m ) $ possible combinations.So far only above three pairs are found to be at all possible with integer values in degrees.The plot has two curves, all combinations are found in the vertical arm. Geometrical meaning of the more horizontal curve is not yet clear. However, it is outside the scope of question OP asks, I attempted a generalization.


An admittedly more direct/simpler way following OP's procedure gets us to :

$$ 2 \cos \theta \sin (\alpha - \theta) - \sin ( 2 \alpha - \theta) =0 $$

A solution in fourth order trig equation $ \theta = f( \alpha) $ may be possible in closed form. Else may be left in the implicit form for numerical iteration. Its Contour graph is also included. The matter why they appear different (crossing vs. non-crossing) still needs attention.

  • $\begingroup$ You've ignored the constraint that $AE = BE$. $\endgroup$ – Anon Sep 9 '16 at 8:52
  • $\begingroup$ Indeed. Thanks for pointing out. $\endgroup$ – Narasimham Sep 9 '16 at 11:02

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