Question

In the figure, seg DE || side AC andseg EF || side BA then prove that AD/DB × AF/FC =1​

Answer 1

Given:

ABC is a triangle,

where, DE || AC and EF || BA.

To find:

$\sf\large\ \frac{AD}{DB} × \frac{AF}{FC}$=1

Concept:

☞The question is completely based on Basic proportionality theorem.

☞Statement: A line parallel to one side of a triangle divides the other two sides into parts of equal proportion.

☞Formula: $\sf\large\ \frac{AB}{BC} × \frac{AE}{EC}$

Solution:

$\sf\ In\:ABC , DE || AC$

•°• $\sf\large\ \frac{BE}{EC}=\frac{BD}{DA} \dashrightarrow\: (1)$

$\sf\ In \:ABC , EF || AB$

•°•$\sf\large\ \frac{CE}{EB}=\frac{CF}{FA} \dashrightarrow\: (2)$

Multiplying (1) and (2), we get:

$\dashrightarrow\: \sf\large\ \frac{BE}{EC}×\frac{CF}{EB}=\frac{BD}{DA}×\frac{CF}{FA}$

$\dashrightarrow\: \sf\ 1= \large\ \frac{DB}{AD}=\frac{FC}{AF}$

$\dashrightarrow\: \underline{\boxed{\bf{\orange{\frac{AD}{DB}=\frac{AF}{FC} =1}}}}$