We are given angles A and B (70 and 60 respectively). Also AΓ=ΒΔ. Ζ and E are midpoints of AB and ΓΔ respectively.

I also drew some bigger circles with radius AH and ΒΘ, trying to see some pattern but with no luck. Geogebra shows that the required angle is 95 degrees but I don't have any clue on how to prove it :( Obviously all midpoints of the 4th side of the quadrilateral, lie on the same line but I can't find its properties.

Geometry is not my strong point!

enter image description here

...and here is the original shape:

enter image description here

  • $\begingroup$ Could you please share the original question without any added constructions? $\endgroup$
    – User1234
    Commented Jun 17, 2018 at 13:41
  • $\begingroup$ @Ishan: Here you are. $\endgroup$
    – Samuel
    Commented Jun 17, 2018 at 13:59

2 Answers 2


Let us shift this problem to the Coordinate Plane with $A$ at the origin, $AB$ on the $X$ axis, and $AB=2\alpha$. Thus $A\equiv(0,0)$ and $B\equiv(2\alpha,0)$. Also $A\Gamma=B\Delta=r$. Lastly let $\theta=\angle EZB$ $$$$ The equation of $A\Gamma$ is $y=x\tan70^{\circ}$ and the equation of $B\Delta$ is $y=x\tan120^{\circ}-2\alpha\tan120^{\circ}$. $$$$ Thus $\Gamma\equiv (x_1,x_1\tan70^{\circ})$ and $\Delta\equiv (x_2,x_2\tan120^{\circ}-2\alpha\tan120^{\circ})$. $$$$ Thus the midpoint $E$ has coordinates $\left(\dfrac{x_1+x_2}2, \dfrac{x_1\tan70^{\circ}+x_2\tan120^{\circ}-2\alpha\tan120^{\circ}}2\right)$.

$$$$ Also $$r=x_1\sec70^{\circ}\Rightarrow x_1=r\cos70^{\circ}$$ and $$r=x_2\sec120^{\circ}-2\alpha\sec120^{\circ}\Rightarrow x_2=(r+2\alpha\sec120^{\circ})\cos120^{\circ}$$ $$$$

Replacing $x_1,x_2$ with their equivalent expressions (this looks messy at first but everything gets simplified), $E$ has the coordinates $$$$ $$\left(\dfrac{r\cos70^{\circ}+(r+2\alpha\sec120^{\circ})\cos120^{\circ}}2,\dfrac{{r\cos70^{\circ}\tan70^{\circ}+(r+2\alpha\sec120^{\circ})\cos120^{\circ}\tan120^{\circ}-2\alpha\tan120^{\circ}}}2\right)$$ $$$$ On simplifying, $$E\equiv \left(\dfrac{r\cos70^{\circ}+r\cos120^{\circ}+2\alpha}2,\dfrac{r\sin70^{\circ}+r\sin120^{\circ}}2\right)$$ $$$$ Lastly since $Z$ is the midpoint of $AB$, $Z$ has the coordinates $(\alpha,0)$.

$$$$Thus, the slope of $EZ$ is $$\tan\theta=\left( \dfrac{\dfrac{r\sin70^{\circ}+r\sin120^{\circ}}2-0}{\dfrac{r\cos70^{\circ}+r\cos120^{\circ}+2\alpha}2-\alpha}\right)$$ $$$$ $$\Rightarrow\tan\theta=\left( \dfrac{{r\sin70^{\circ}+r\sin120^{\circ}}}{{r\cos70^{\circ}+r\cos120^{\circ}}}\right)$$ $$$$

$$\Rightarrow\theta=\tan^{-1}\left( \dfrac{{r\sin70^{\circ}+r\sin120^{\circ}}}{{r\cos70^{\circ}+r\cos120^{\circ}}}\right)$$ $$$$

$$=\tan^{-1}\left( \dfrac{{\sin70^{\circ}+\sin120^{\circ}}}{{\cos70^{\circ}+\cos120^{\circ}}}\right)$$ $$$$

$$=\tan^{-1}\left( \dfrac{{2\sin95^{\circ}\cos25^{\circ}}}{{2\cos95^{\circ}\cos25^{\circ}}}\right)=\tan^{-1}(\tan95^{\circ})$$ $$$$ Thus $\theta=95^{\circ}$

  • $\begingroup$ "the median though P on AB is also the internal angle bisector". Why? Also Z, E, I and P are not collinear. $\endgroup$
    – Samuel
    Commented Jun 17, 2018 at 15:24
  • $\begingroup$ OK, Thank you! I haven't checked your trigonometry solution, but I prefer geometry in any case! $\endgroup$
    – Samuel
    Commented Jun 17, 2018 at 15:28
  • $\begingroup$ Also, if we extend EZ it will not pass through P. $\endgroup$
    – Samuel
    Commented Jun 17, 2018 at 15:32
  • $\begingroup$ No problem, I will go through it. Thank you $\endgroup$
    – Samuel
    Commented Jun 17, 2018 at 16:00
  • $\begingroup$ You're welcome! Hope the solution helps. $\endgroup$
    – User1234
    Commented Jun 17, 2018 at 16:06

Alt. hint:  define a complex plane with $\,Z=0, B=1, A=-1, A\Gamma=B\Delta=\lambda \in \mathbb{R}^+\,$.

Let $\,\alpha=70^\circ=\frac{7 \pi}{18}\,$, $\,\beta=60^\circ=\frac{\pi}{3}\,$, then $\,\Gamma = -1 + \lambda e^{i\alpha}\,$ and $\,\Delta = 1 + \lambda e^{i(\pi - \beta)}\,$.

The midpoint of $\Gamma\Delta$ is $\require{cancel}\,E=\frac{1}{2}(\Gamma+\Delta)=\frac{\lambda}{2}\left(\cancel{-1}+ e^{i\alpha}+\cancel{1}+e^{i (\pi-\beta)}\right)=\frac{\lambda}{2}\left( e^{i\,7\pi/18}+e^{i \,2 \pi / 3}\right)\,$.

Angle $\,\angle EZB = \arg E\,$, then using the identities for sums of sines/cosines:

$$ \begin{align} \frac{2}{\lambda}\,E &\,=\, \left(\cos 70^\circ+\cos 120^\circ\right) + i \left(\sin 70^\circ+\sin 120^\circ\right) \\ &\,=\, 2 \cos \frac{70^\circ+120^\circ}{2} \cos \frac{120^\circ - 70^\circ}{2} + 2i \sin \frac{70^\circ+120^\circ}{2} \cos \frac{120^\circ - 70^\circ}{2} \\ &\,=\, 2 \cos 25^\circ \left( \cos 95^\circ + i \sin 95^\circ\right) \end{align} $$


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