Thales Application

Question

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

Answer 1

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 $CF$

Answer 2

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

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