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 #70211 – 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.

$$\overrightarrow{EF} \parallel \overrightarrow{BD} \Rightarrow \Delta AFE \sim \Delta ABD$$

Let

$$\frac{FE}{BD} = \frac{AF}{AB} = \frac{AE}{AD} = k$$ $$\overrightarrow{ER} \parallel \overrightarrow{CD} \Rightarrow \Delta AGE \sim \Delta ACD$$

Let

$$\frac{EG}{DC} = \frac{AG}{AC} = \frac{AE}{AD} = m$$

Then $m = k$

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

Then

$$\overrightarrow{FG} = \overrightarrow{EG} - \overrightarrow{EF}$$ $$\overrightarrow{BC} = \overrightarrow{DC} - \overrightarrow{DB}$$ $$\overrightarrow{DB} = \frac{1}{k} \overrightarrow{EF}$$ $$\overrightarrow{DC} = \frac{1}{m} \overrightarrow{EG} = \frac{1}{k} \overrightarrow{EG}$$ $$\overrightarrow {B C} = \overrightarrow {D C} - \overrightarrow {D B} = \frac {1}{k} \overrightarrow {E G} - \frac {1}{k} \overrightarrow {E F} = \frac {1}{k} \left(\overrightarrow {E G} - \overrightarrow {E F}\right) = \frac {1}{k} \overrightarrow {F G}$$

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

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