State and prove mid-point theorum and write the converse of mid-point theorum and prove it.
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theorem1 .the straight lines joining mid points of any two sides of a triangle is parallel to the third side and is equal to half of it .
given. a triangle ABC, DandE are midpoints of AB and AC respectively.
RTP. DE||BC and DE=1/2BC
construction. through C, draw a st. line parallel to BA to meet DE produced at F
proof . in triangle ADE AND ECF
1. AE=EC. (E IS THE MID POINT OF AC)
2. <DEA=<CEF ( VERTICALLY OPPOSITE ANGLE)
3. <DAE=<ECF. (ALTERNATE ANGLES SINCE BA||CF)
∆ADE~ ∆ECF (ASA CRITERIA)
DE = CF AND AD = CF BY CPCT -5
AD=BD. (D IS THE MID POINT OF AB ) GIVEN-6
BD=CF (FROM 5 AND 6)
THEN DBCF IS A PARALLELOGRAM AND DE||BC AND DE=BC
DE=1/2DF FROM-5
DE=1/2BC. SINCE DF=BC
HENCE DE || BC AND DE=1/2 BC
THIS IS MID POINT THEOREM
CONVERSE OF THIS THEOREM IS
THE ST LINE DRAWN THROUGH THE MID POINT OF ONE SIDE OF A TRIANGLE IS PARALLEL TO ANOTHER SIDE BISECTS THE THIRD SIDE
HERE YOU HAVE TO PROVE AE= EC
BY THE HELP OF AAS CRITERIA AE=EC BY CPCT
HOPE THIS WILL HELP YOU
given. a triangle ABC, DandE are midpoints of AB and AC respectively.
RTP. DE||BC and DE=1/2BC
construction. through C, draw a st. line parallel to BA to meet DE produced at F
proof . in triangle ADE AND ECF
1. AE=EC. (E IS THE MID POINT OF AC)
2. <DEA=<CEF ( VERTICALLY OPPOSITE ANGLE)
3. <DAE=<ECF. (ALTERNATE ANGLES SINCE BA||CF)
∆ADE~ ∆ECF (ASA CRITERIA)
DE = CF AND AD = CF BY CPCT -5
AD=BD. (D IS THE MID POINT OF AB ) GIVEN-6
BD=CF (FROM 5 AND 6)
THEN DBCF IS A PARALLELOGRAM AND DE||BC AND DE=BC
DE=1/2DF FROM-5
DE=1/2BC. SINCE DF=BC
HENCE DE || BC AND DE=1/2 BC
THIS IS MID POINT THEOREM
CONVERSE OF THIS THEOREM IS
THE ST LINE DRAWN THROUGH THE MID POINT OF ONE SIDE OF A TRIANGLE IS PARALLEL TO ANOTHER SIDE BISECTS THE THIRD SIDE
HERE YOU HAVE TO PROVE AE= EC
BY THE HELP OF AAS CRITERIA AE=EC BY CPCT
HOPE THIS WILL HELP YOU
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