Question

Through the vertex $D$ of a parallelogram $ABCD,$ a line is drawn is the sides $BA$ and $BC$ produced at $E$ and $F$ respectively then $\frac{DA}{AE}=\frac{FB}{BE}=\frac{FC}{CD}$

• A. True
• B. False

Answer 1

The correct option is A True

In $△DCF$ and $△EAD$

$\angle 2=\angle 1$(corresponding angles)

and $\angle 4=\angle 3$(corresponding angles)

$\therefore$ by using $AA$ similar condition,

$△DCF\sim △EAD$

$\Rightarrow \frac{DC}{EA}=\frac{CF}{AD}=\frac{DF}{ED}$

$\Rightarrow \frac{DC}{EA}=\frac{CF}{AD}$

$\Rightarrow \frac{DC}{CF}=\frac{EA}{AD}$

$\Rightarrow \frac{CF}{DC}=\frac{AD}{EA}$ .......$\left(1\right)$

Now, in $△EAD$ and $△EBF$

$\angle 1=\angle 1$(common)

and $\angle 3=\angle 4$(corresponding angles)

$\therefore$ by using $AA$ similar condition

$△EAD\sim △EBF$

$\Rightarrow \frac{EA}{EB}=\frac{AD}{BF}=\frac{ED}{EF}$

$\Rightarrow \frac{EA}{EB}=\frac{AD}{BF}$

$\Rightarrow \frac{AD}{AE}=\frac{BF}{BE}$ .......$\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$, we get

$\frac{DA}{AE}=\frac{FB}{BE}=\frac{FC}{CD}$

Hence the given statement is true.