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I'm struggling with this exercice:

Let $ABCD$ be a trapezium. $AB$ and $CD$ intersect at $O$.

$OE \parallel BD$, $OF \parallel AC $ and $F,C,B,E$ are aligned points

  • Prove that $EB=CF$

I think it's an application of Thales' theorem. I applied the theorem to the $\triangle$ s $OBF,OCE,OBC$ , but I don't see how to use it to prove that $EB=CF$.

I also noticed that we can prove that the triangles $OBE$ and $OCF$ are equal, $OE=OF$ , $OB=OC$, but I still need an equal angle in both triangles. Thanks

Sinπ
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Rfgauss
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  • Hi and welcome to MSE! Please use MathJax and provide an image (I would recommend you GeoGebra). This might help other users when it comes to answering the question. In my case, for instance, I find it difficult to understand what the exercise is about and what conditions are given... An image, as said, might also be very helpful ;) – Dr. Mathva Apr 22 '19 at 18:12
  • @mathva I just did, thanks – Rfgauss Apr 22 '19 at 22:22

2 Answers2

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You are right ! The problem does indeed involve Thales' theorem , and similarity:-

Note that we have:-

$\triangle OEC \sim \triangle DBC \implies \frac{EC}{BC}=\frac{OC}{DC} \tag{1}$ $\triangle OFB \sim \triangle ACB \implies \frac{BF}{BC}=\frac{OB}{AB} \tag{2}$

But by Thales' Theorem , the RHS of $(1)$ and $(2)$ are equal !

Therefore , $EC$ equals $BF$ , which implies $EB$ equals $CF$

Sinπ
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1

Just Thales 3 times, in angles $\angle ECO, \angle BOC$ and $\angle FBO$:

$$\color{red}{EB\over BC} = {OD\over DC} = {OA\over AB} = \color{red}{CF\over BC}\implies EB = CF$$

nonuser
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