Question

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Answer 1

Answer: Proved below

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Answer 2

Answer:

Internal angle bisector theorem

In ∆AOB ---> (I)

$\frac{AO}{BO} = \frac{AF}{BF}$

In∆BOC --->(Il)

$\frac{BO}{OC } = \frac{BD}{CD}$

In∆COA --->(III)

$\frac{CO}{CE} = \frac{CE}{AE}$

Multiplying (I),(II) and (III) , we get

$\frac{AO}{BO} \times \frac{BO}{CO} \times \frac{CO}{AO} = \frac{AF}{BF} \times \frac{BD}{CD} \times \frac{CE}{AE}$

$Therefore, \: AF \times BA \times CE \times BF \times CD \times AE$

I Hope It Will Help!

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