Question

plz Answer it with a diagram​

Answer 1

Answer:

The diagram is in attachment!

Step-by-step explanation:

In ∆BDF and ∆BCE,

$\angle{DBF} = \angle{CBE} \: \: \: \: \: \: (BF \: is \: the \: angle \: bisector \: of \: \angle \: DBC)$

$\angle{DBF} = \angle{CBE}$

( BD and AC are angle bisectors)

so, ∆BDF ~ ∆ BCE (by AA similarity criterion)

$= > \frac{BF}{DF} = \frac{BE}{CE}$

=> BF × CE = BE × DF.

Hence, Proved!

Answer 2

Step-by-step explanation:

in ∆BDF AND ∆BCD

DBE =

CBE ( CF IS BISECTAR and angal 1)

DBE-CBE

BD and AC are the BISECTAR

so ∆ BDF ~∆ BCE ( by AA is simmilarty )

=>BF/ DE , BE /CF

=> BE× CE× BE×DF

HENCE PROVe:-

I hope it helps for all tq khudahafiz