Question

ABCD is a plane quadrilateral and E is any point on AD. EF is drawn parallel to DB to meet AB in F and EG is drawn parallel to DC to meet AC in G. Prove that FG is parallel to BC.

Answer 1

ANSWER on Question #81271 – Math – Geometry

QUESTION

$ABCD$ is a plane quadrilateral and $E$ is any point on $AD$. $EF$ is drawn parallel to $DB$ to meet $AB$ in $F$ and $EG$ is drawn parallel to $DC$ to meet $AC$ in $G$. Prove that $FG$ is parallel to $BC$.

SOLUTION

$$EF \parallel BD \rightarrow \Delta AFE \sim \Delta ABD \rightarrow \frac{AF}{AB} = \frac{FE}{BD} = \frac{AE}{AD} = k$$ $$ER \parallel CD \rightarrow \Delta AGE \sim \Delta ACD \rightarrow \frac{AG}{AC} = \frac{GE}{CD} = \frac{AE}{AD} = m$$

Then,

$$k = \frac{AE}{AD} = m \rightarrow \boxed{m = k}$$

We have vectors $\overrightarrow{EF}, \overrightarrow{DB}, \overrightarrow{EG}, \overrightarrow{DC}, \overrightarrow{FG}$, and $\overrightarrow{BC}$.

$$\overrightarrow{FG} = \overrightarrow{EG} - \overrightarrow{EF}$$ $$\overrightarrow{BC} = \overrightarrow{DC} - \overrightarrow{DB}$$ $$\overrightarrow{DB} = \frac{1}{k} \cdot \overrightarrow{EF}$$ $$\overrightarrow{DC} = \frac{1}{m} \cdot \overrightarrow{EG} = \frac{1}{k} \cdot \overrightarrow{EG}$$

Then,

$$\overrightarrow{BC} = \overrightarrow{DC} - \overrightarrow{DB} = \frac{1}{k} \cdot \overrightarrow{EG} - \frac{1}{k} \cdot \overrightarrow{EF} = \frac{1}{k} \cdot (\overrightarrow{EG} - \overrightarrow{EF}) = \frac{1}{k} \cdot \overrightarrow{FG}$$ $$\boxed{\overrightarrow{BC} = \frac{1}{k} \cdot \overrightarrow{FG}}$$

Conclusion,

The vectors $\overrightarrow{BC}$ and $\overrightarrow{FG}$ are collinear vectors. Hence, $BC \parallel FG$

Q.E.D.

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