explain mid point theorem with proof
Answers
Answer:
Mid - Point Theorem
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 triangle.”
Proof:
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 sides BC, whereas the side DE is half of the side BC; i.e.
DE is parallel to BC
DE = (1/2*BC).
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)
angle CEF = angle AED {vertically opposite angles}
EF = DE { by construction}
hence,
triangle CFE is congruent to triangle ADE {by SAS}
Therefore,
angle CFE = angle ADE {by c.p.c.t.}
angle FCE = angle DAE {by c.p.c.t.}
and CF = AD {by c.p.c.t.}
The angles, angle CFE and angle ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.
In a similar way, angle FCE and angle DAE are the alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.
Therefore, CF is parallel to AB
So, CF is parallel to 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 is parallel to DF
and BC = DF
BC is parallel to DE
and DE = (1/2 * BC).
Hence, the midpoint theorem is Proved.
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.
