Question

state and prove converse of mid point theorem​

Answer 1

converse of mid-point theorem: it states that in a triangle line drawn from the mid-point of the one side of triangle, parallel to the other side intersect the third side at its mid-point. ... from D a line DE is drawn parallel to BC, intersect AC at E.
MidPoint Theorem Proof
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.

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).

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 brainlist