Question

The diagonal of BD of a parallelogram ABCD intersects the segment AE at the point F,where E is the point on the side BC.Prove that DF×EF=FB×FA.

Answer 1

SOLUTION

Consider ∆AFD & ∆BFE

Then, we have

=)∠1= ∠2

=)∠3= ∠4

Therefore, AA- criterion of similarity, we have

=) ∆FBE~ ∆FDA

$= > \frac{</strong><strong>FB</strong><strong>}{</strong><strong>FD</strong><strong>} = \frac{</strong><strong>FE</strong><strong>}{</strong><strong>FA</strong><strong>} \\ \\ = > \frac{</strong><strong>FB</strong><strong>}{</strong><strong>DF</strong><strong>} = \frac{</strong><strong>EF</strong><strong>}{</strong><strong>FA</strong><strong>}$

Hence, DF× EF= FB× FA [proved]

hope it helps ☺️

Answer 2

Answer:

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