State and prove the mid-point theorem
Answers
Heya,
STATEMENT:
Line segments joining the midpoints of
sides of a triangle is parallel to the third
side and half of third side.
GIVEN:
∆ABC in which E & Fare the mid points
of AB and AC
TO PROVE:
EFIl BC and EF=1/2BC
CONSTRUCTION:
Extend EF to the point D such that CDII
BA
PROOF:
∆AEF and ∆FCD
angle 1= 2 (alternate interior angle]=>A
AF=FC [F is the mid point]=>S
angle 3= 4 [Vertically oppositangles]=>A
Therefore,
∆AEF ~= ∆FCD[ASA ]
EF = FD , AE = CD [ CPCT ]
BE=AE=CD
=> BE=CD, BE||CD
BECD is a parallelogram
=> ED=B, ED BC
=> EF+FD=BC, EFI|BC
=> EF+EF=BC, EFIBC
=> 2EF=BC, EF||BC
=> EF= 1/2BC, EF BC
Hope my answer helps you
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MidPoint Theorem Statement
The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”
Construction- Extend the line segment DE and produce it to F such that, EF=DE.
In the triangle, ADE, and also the triangle CFE
EC= AE —– (given)
∠CEF = ∠AED {vertically opposite angles}
EF = DE { by construction}
hence,
△ CFE ≅ △ ADE {by SAS}
Therefore,
∠CFE = ∠ADE {by c.p.c.t.}
∠FCE= ∠DAE {by c.p.c.t.}
and CF = AD {by c.p.c.t.}
The angles, ∠CFE and ∠ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.
In a similar way, ∠FCE and ∠DAE are the alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.
Therefore, CF ∥ AB
So, CF ∥ BD
and CF = BD {since BD = AD, it is proved that CF = AD}
Thus, BDFC forms a parallelogram.
By the use of properties of a parallelogram, we can write
BC ∥ DF
and BC = DF
BC ∥ DE
and DE = (1/2 * BC).
Hence, the midpoint theorem is Proved.
