Let $ABCD$ be a trapezoid with $AB \parallel CD$. Then let $\frac{AE}{AD} = \frac{BF}{BC}$ . Prove $EF$ parallel to $AB$

Question

Let $ABCD$ be a trapezoid with $AB \parallel CD$. Let line $E$ and $F$ be points on $\overline{AD}$ and $\overline{BC}$ respectively such that $\frac{AE}{AD} = \frac{BF}{BC}$ Prove $\overleftrightarrow{EF} \parallel \overleftrightarrow{AB}$

1. Extend the segments $AD$ and $BC$ to meet at a point $P$.
2. Now suppose you have another point $F'$ on $\overline{BC}$ such that $\overleftrightarrow{EF'}\parallel \overleftrightarrow{AB}$. Then $\overleftrightarrow{EF'} \parallel \overleftrightarrow{DC}$ by transitivity.
3. in $\triangle ABP$ by thales theorem we have that $\frac{PD}{DA}=\frac{PC}{CB} \rightarrow \frac{PD}{PC}=\frac{DA}{CB}$
4. In $\triangle DEF'$ by thales theorem we have that $\frac{PD}{DE}=\frac{PC}{CF'} \rightarrow \frac{PD}{PC}=\frac{DE}{CF'}$
5. Thus combining the results from above we have that $\frac{PD}{PC}=\frac{DA}{CB}=\frac{DE}{CF'}$. Therefore we have that $\frac{AD}{DE}=\frac{BC}{CF'}\rightarrow \frac{DE}{AD}=\frac{CF'}{BC}$ Now subtracting $1$ from both sides of the equation we get that $\frac{AD-DE}{AD}=\frac{BC-CF'}{BC} \rightarrow \frac{AE}{AD}=\frac{BF'}{BC}$

6. Now given that $\frac{AE}{AD}=\frac{BF}{BC}$ then $\frac{BF'}{BC}=\frac{BF}{BC}$ which implies that $F'$ coincides with the point $F$.

7. Therefore given that $\overleftrightarrow{EF'} \parallel \overleftrightarrow{AB}$ then $\overleftrightarrow{EF} \parallel \overleftrightarrow{AB}$

Looking for other approaches as well since I know with geometry in particular you can get a lot of different solution to a problem.

Answer 1

Extend $AD$ and $BC$ to meet at a point $P$. Then $\Delta PAB\sim \Delta PDC$ and hence $$ \frac{PA}{AD}=\frac{PB}{BC}. \tag{1}$$ On the other hand, $$ \frac{AE}{AD} = \frac{BF}{BC} \tag{2}$$ from which one has $$ \frac{AD}{BC} = \frac{AE}{BF} \tag{3}.$$ Adding (1) to (2), one has $$ \frac{PE}{AD} = \frac{PF}{BC} $$ or $$ \frac{PE}{PF} = \frac{AD}{BC}. \tag{4}$$ From (3) and (4), one has $$ \frac{PE}{PF}=\frac{AE}{BF}$$ or $$ \frac{PE}{AE}=\frac{PF}{BF}$$ which implies $\Delta PAB\sim \Delta PEF$. So $\angle PAB=\angle PEF$ and hence $EF//AB$.

Answer 2

The following images prove the statement.