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MATHS PROBLEM HELP. BEST ANSWER GETS 10 POINTS?
In a parallelogram ABCD, the diagnal BD intersects the segment AE at F. Where E is any point on side BC.
Prove that DFxEF=FBxFA
1 Answer
Relevance
- Let'squestionLv 77 years ago
We have two similar triangles: BFE and DFA because angle F = angle D vertically opposite angles and
angle B = angle D and angle E = angle A being pair of alternate angles.
So e/a = b/d or (FB/DF) = (EF/FA) or
DF*EF = FB*FA
Hence proved.
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