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In Parallelogram $ABCD$, $|BD|=|AB|$. The point $F$ on the side $CD$ is chosen such that $|BF|=|BC|=|FD|$. Find the angle $\angle ABC.$

I've tried using the law of sines and cosines but to no avail. I get way too many variables to deal with. I don't think It should be that hard.

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Kindly consider this rough figure and assume that $\angle A=\theta$ with BF=BC=DF and AB=BD.

Since AB=BD, then $\angle ADB=\theta$ ( isosceles triangle) and $\angle C=\pi-\theta$ (Parallelogram's opposite vertices have supplementary angles).

Now since BF=BC, $\angle C=\theta=\angle BFC$ (isosceles triangles again) making $\angle FBC=2\theta-\pi$

Now, $\angle DFB=\theta$ (Linear angle with $\angle CFB$)

also BF=DF making $\angle FDB=\frac{\pi-\theta}{2}=\angle FBD$

Now, $\angle ABD=\angle FBD$

$$\frac{\pi-\theta}{2}=\pi-2\theta$$ $$\theta=\frac{\pi}{3}$$

Thus $\angle ABC=\frac{\pi-\theta}{2}=\frac{\pi}{3}=60^o$

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A simple solution based on the angle.

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It remains to solve the equation and obtain an angle $ \angle \alpha $.

Do not forget that $ \angle ABC = 180 - \angle \alpha $.

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    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig May 7 '17 at 13:07
  • $\begingroup$ Thanks. But I know how to do this. It was convenient to write on paper because i'm of the road $\endgroup$ – Dmitry May 7 '17 at 13:16

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