Question

in fig. seg DE || side AC and seg EF || side BA then prove that AD\DB×AF/FC=1​

Answer 1

Step-by-step explanation:

Hope it helps you friend.

Answer 2

Answer with Step-by-step explanation:

DE is parallel to AC

EF is parallel to side AB

In triangle ABC

DE is parallel to AC

By triangle proportionality theorem

$\frac{BE}{EC}=\frac{BD}{DA}$..(1)

EF is parallel to AB

By triangle proportionality theorem

$\frac{BE}{EC}=\frac{AF}{FC}$..(2)

Divide equation (2) by (1)

$1=\frac{\frac{AF}{FC}}{\frac{BD}{DA}}$

$1=\frac{AF}{FC}\times \frac{AD}{DB}$

Hence, proved.

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