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Question^^^
AD // BC , AC Intersect BD = M
E is the midpoint of BC (ME IS MEDIAN)
Provte that
1) Area of Triangle ABM = area of triangle DCM
2) Area of the figure ABEM = Area of the figure DCEM.
My proof: (Please let me know if there is any mistakes) The model answer says another solution which i don't understand..
Proof:
Since Triangle BAD and Triangle CAD have common base DA and AD // BC
Therefore, Triangle BAD = Triangle CAD in area. (1)
-Subtracting Triangle DMA
Therefore, Triangle ABM = Triangle DCM In area (First required) (2)
In Triangle MBC
Since E is midpoint of BC , EC=BE
Therefore, ME is a median
Therefore, Triangle MEB = Triangle MEC In area (3)
Adding (2) and (3)
we deduce that:
Triangle ABM + Triangle MEB = Figure ABEM = The area of Triangle DCM + Triangle MEC = The figure DCEM
Therefore, the figure ABEM = The area of the figure DCEM.
Ok so that was my proof..
Here is the model answer..
:
∵ ∆∆ ABC , DBC have a common base BC , BC ∕∕ AD ∴ the area of ∆ ABC = the area of ∆ DBC ( 1 ) In ∆ MBC , ∵ ME is a median ∴ the area of ∆ MEC = the area of MEB ( 2 ) Subtracting ( 2 ) from ( 1 ) ∴ the area of the figure ABEM = the area of the figure DMEC
Why did he subtract and is my solution considered to be correct?
