what is mid point thereom prove it?
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statement :-
"In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it."
proof:-
Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined.
Given: AD = DB and AE = EC.
To Prove: DE ∥ BC and DE = 1212 BC.
Construction: Extend line segment DE to F such that DE = EF.
Proof: In △△ ADE and △△ CFE
AE = EC (given)
∠∠AED = ∠∠CEF (vertically opposite angles)
DE = EF (construction)
hence
△△ ADE ≅≅ △△ CFE (by SAS)
Therefore,
∠∠ADE = ∠∠CFE (by c.p.c.t.)
∠∠DAE = ∠∠FCE (by c.p.c.t.)
and AD = CF (by c.p.c.t.)
The angles ∠∠ADE and ∠∠CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF.
Similarly, ∠∠DAE and ∠∠FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC.
Therefore, AB ∥ CF
So, BD ∥ CF
and BD = CF (since AD = BD and it is proved above that AD = CF)
Thus, BDFC is a parallelogram.
By the properties of parallelogram, we have
DF ∥ BC
and DF = BC
DE ∥ BC
and DE = 1212BC (DE = EF by construction)
Hence proved.
"In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it."
proof:-
Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined.
Given: AD = DB and AE = EC.
To Prove: DE ∥ BC and DE = 1212 BC.
Construction: Extend line segment DE to F such that DE = EF.
Proof: In △△ ADE and △△ CFE
AE = EC (given)
∠∠AED = ∠∠CEF (vertically opposite angles)
DE = EF (construction)
hence
△△ ADE ≅≅ △△ CFE (by SAS)
Therefore,
∠∠ADE = ∠∠CFE (by c.p.c.t.)
∠∠DAE = ∠∠FCE (by c.p.c.t.)
and AD = CF (by c.p.c.t.)
The angles ∠∠ADE and ∠∠CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF.
Similarly, ∠∠DAE and ∠∠FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC.
Therefore, AB ∥ CF
So, BD ∥ CF
and BD = CF (since AD = BD and it is proved above that AD = CF)
Thus, BDFC is a parallelogram.
By the properties of parallelogram, we have
DF ∥ BC
and DF = BC
DE ∥ BC
and DE = 1212BC (DE = EF by construction)
Hence proved.
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