Question

What is midpoint theorem proof

Answer 1

Answer:

Construction-  Extend the line segment DE and produce it to F such that, EF = DE.

In triangle ADE and CFE,

EC = AE —–   (given)

∠CEF = ∠AED (vertically opposite angles)

EF = DE (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.

please mark me as the brainliest

Answer 2

If the line segment adjoins midpoints of any of the sides of a triangle, then the line segment is said to be parallel to all the remaining sides, and it measures about half of the remaining sides.Let E and D be the midpoints of the sides AC and AB.