Question

in figure AC||DE and AE||DF .Prove that BE2=BF×BC?​

Answer 1

Answer:

In △ABC,

DE∥AC

∴ By proportionality theorem,

DA/BD= EC/BE ............(1)

Similarly,

In △ABE,

DF∥AE

∴ By proportionality theorem,

DA/BD = FE/BF ............(2)

EC/BE= FE/BF.

Hence Proved.

Answer 2

ANSWER:

basic porpotionlaly therom

(BPT) : if a line is a draw parallel two one side of triangle to intersect the other two other sides in distnict point then the otherr two sides are divided in the smae ratio that is also know thales therom

SOLUTIONS:

given in triangle BAC,DE||AC

BE/EC=BD/DA………………(1)

(by Thales therom)

in triangle BAE ,DF|| AE is given

BF/FE=BD/DA………………(2)

(by Thales therom) BPT

from eq 1and2

BF /FE=BE/EC

FB/BF=EC/BF

$reciporcal \: the \: term$

$hence \: provided \: it$