Question

prove mid point theorem

Answer 1

consider a triangle abc with mid point of AB labelled m now construction of line parallel to m

this theorem states that two sides of the triangle is parallel to the third side

Answer 2

Here, In △ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined.

Given: AD = DB and AE = EC.

To Prove: DE ∥ BC and DE = 1/2 BC.

Construction: Extend line segment DE to F such that DE = EF.

Proof: In △ ADE and △ CFE

AE = EC   (given)

∠AED = ∠CEF (vertically opposite angles)

DE = EF   (construction)

hence

△ ADE ≅ △ CFE (by SAS)

Therefore,
∠ADE = ∠CFE   (by c.p.c.t.)

∠DAE = ∠FCE   (by c.p.c.t.)

and AD = CF  (by c.p.c.t.)

The angles ∠ADE and ∠CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF.

Similarly, ∠DAE and ∠FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC.

Therefore, AB ∥ CF

So, BD ∥ CF

and BD = CF (since AD = BD and it is proved above that AD = CF)

Thus, BDFC is a parallelogram.

By the properties of parallelogram, we have

DF ∥ BC

and DF = BC

DE ∥ BC

and DE = 1/2 BC  (DE = EF by construction)

Hence proved.