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My old high school teacher has been posting some math problems online and I just couldn't solve this one. Let triangles $\triangle ABC$ and $\triangle BDE$ be equilateral. Prove $$\overline{FB}=\sqrt{\overline{CF}\cdot \overline{FE}}$$

This seems just plain awful using analytic geometry, but it is doable, because we know the coordinates of every point taking $A$ as the origin and $AD$ as the x-axis and the y-axis perpendicular to it and pointing upwards. $C=(l_1/2,l_1\sqrt{3}/2)$, for example, so it is just a matter of doing computations.

This former teacher of mine usually uses clever tricks, but I can't seem to think of any.

Kadmos
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    Hints: (a) What you need to prove is equivalent to $|CF|:|FB|=|FB|:|FE|$ so you probably just want to prove $\triangle CFB\sim\triangle BFE$. (b) By rotating $\triangle ABE$ around $B$ clockwise by $60^\circ$ you get $\triangle CBD$; (c) The previous implies that $\angle BAF=\angle BCF$ and the quadrilateral $ABFC$ is cyclic. Similarly quadrilateral $BDEF$ is cyclic. Now you can use the circle theorems and angle chasing to prove that some angles are equal and thereby prove the similarity. –  Jun 27 '23 at 12:39
  • @StinkingBishop, very clever argument. Thank you! But how did you think of showing $\triangle ABE \approx \triangle CBD$ so quickly? Is this a standard proof one should know? – Kadmos Jun 27 '23 at 12:50
  • How standard - I don't know. I've certainly seen things like that before. –  Jun 27 '23 at 12:55
  • Btw, if you write your comment verbatim as an aswer I will accept it as the best answer. Loved your cyclic quadrilaterals. – Kadmos Jun 27 '23 at 12:56
  • I will see how much time I have. It would be nice to add a few pictures etc. Also I don't mind if someone comes up with the same (or a different) proof in the meantime. –  Jun 27 '23 at 12:58
  • I think yours is already a full proof. Lol. – Kadmos Jun 27 '23 at 12:58
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    (I also didn't look for duplicates; if the same question has already been covered on MSE, it is only fair to mark this as a duplicate and link it to the original.) –  Jun 27 '23 at 12:59
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    @Kadmos Recognizing the congruency/rotation is a common trick. For example, this gives us that $|AE| = |CD|$ and they meet at $60^\circ$ immediately, which would be a more common version of this problem. – Calvin Lin Jun 27 '23 at 14:45

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HINT.- If we don't want to use coordinates then we can do as follow.

Because it is easy to get $\overline{AE}^2=\overline{CD}^2=a^2+b^2+ab$ and $\overline{CE}^2=a^2+b^2-ab$ where $a,b$ are the sides of the equilateral triangles, we note $\overline{AE}^2=\overline{CD}^2=c$ and $\overline{CE}^2=d$ and consider as data of the problem $a,b,c,d$. With this we have three unknowns $X,Y,Z$ and we need to verify that $Z^2=XY$ (see at the attached figure).

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Therefore we need three independent equations to determine $X,Y,Z$. We apply Stewart's theorem so we have $$(c-Y)^2b+(c-X)^2a=(a+b)(Z^2+ab)\\d^2(c-Y)+a^2Y=c(X^2+Y(c-Y))\\d^2(c-X)+b^2X=c(Y^2+X(c-X))$$ Solving this system we can verify that $Z^2=XY$.

Piquito
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  • Much nicer than coordinate systems. But I still think @Stinking Bishop's argument is unbeatably elegant... – Kadmos Jun 27 '23 at 19:12
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    @Kadmos I must admit that i could not see any cyclic quadrilateral despite i searched about one. Now i was convinced by a graphic calculator that this is so. – Piquito Jun 27 '23 at 19:55
  • Showing $\angle CAF=\angle CBF$ (or any other pair of similarly defined angles) implies we have a cyclic quadrilateral... Usually this is proven by contradiction: suppouse $B$ doens't lie in the circumference defined by the ponts $C$, $F$ and $A$ and take $B'$ in $CB$ which lies. $\angle CB'F>\angle CBF$ if you think $B$ is outside the circumference, but this is absurd (we know $\angle CB'F=\angle CBF$). Similar contradiction if $B$ is inside the circumference. – Kadmos Jun 27 '23 at 20:13
  • Anyway, I could not localized one. Some years ago i have a good geometrical intuition (change "some years" by "a lot of years". Regards. – Piquito Jun 27 '23 at 20:33
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Accepting the argument in the comment of @StinkingBishop, that quadrilaterals $ABFC$ and $DBFE$ are cyclic, then$$\angle CFB=\angle BFE=120^o$$ Let $BJ$ be the radius of circle $ABFC$.

Since$$\angle EBJ=\angle EBC+\angle CBJ=60^o+30^o=90^o$$then $EB$ is tangent, and$$\angle EBF=\angle BCF$$ TTT Therefore$$\triangle FBC\sim\triangle FEB$$and$$\frac{CF}{FB}=\frac{FB}{FE}$$or$$FB=\sqrt{CF\cdot FE}$$

Edward Porcella
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As shown in figure, we complete the trapezoid BTED. F is intersection of diagonals of parallelograms BTEC and ATQC which are similar. We have the ratio of half digonals as:

$$\frac {EF}{FC}=\frac{FC}{AF}\Rightarrow FC^2=AF\times EF$$

sirous
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