proof the mid point theorem
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Answer:
To prove the midpoint theorem, we use the congruency rules. We construct a triangle outside the given triangle such that it touches the side of the triangle.
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Answer:
If a line segment adjoins the mid-point of any two sides of a triangle, then the line segment is said to be parallel to the remaining third side and its measure will be half of the third side.
Consider the triangle ABC, as shown in the above figure,
Let E and D be the midpoints of the sides AC and AB. Then the line DE is said to be parallel to the side BC, whereas the side DE is half of the side BC; i.e., DE || BC
DE = (1/2 * BC).
Now consider the below figure,
Mid- Point Theorem
Construction- Extend the line segment DE and produce it to F such that, EF = DE and join CF.
In triangle ADE and CFE,
EC = AE —– (given)
∠CEF = ∠AED (vertically opposite angles)
DE = EF ( By construction )
By SAS congruence criterion,
△ CFE ≅ △ ADE
Therefore,
∠CFE = ∠ADE {by c.p.c.t.}
∠FCE= ∠DAE {by c.p.c.t.}
and CF = AD {by c.p.c.t.}
∠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 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.