in a trapezium ABCD,AB||DC and point E is the mid point of the side AD. A line through the point E and parallel to the side AB meet the line BC in F.Prove that F is the mid point of BC?
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ANSWER
ABCD is trapezium in which AB∥DC.
EF is parallel to side DC.
Then we have AB∥DC∥EF.
Hence we have also trapezium ABFE and trapezium EFCD.
Let AP be the perpendicular to DC and this intersects EF at Q.
AQ will be perpendicular to EF.
For △APD and △AQE we have
EA
AD
=
AQ
AP
=2
This gives AP=2AQ
i.e, AQ=QP
Consider the area we have area ABCD= area ABFE+ area EFCD
(
2
1
)AP×(AB+DC)=(
2
1
)AQ×(AB+EF)+(
2
1
)QP×(EF+DC)
⇒AP(AB+DC)=AP×
2
AB
+AP×
2
EF
+AP×
2
EF
+AP×
2
DC
⇒AP×
2
AB
+AP×
2
DC
=AP×
2
EF
+AP×
2
EF
⇒AP×
2
(AB+DC)
=AP×EF
EF=
2
(AB+DC)
Answered by
1
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