Question

prove mid point theorem

Answer 1

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 = 1212 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 = 1212BC  (DE = EF by construction)

Hence proved.

Answer 2

Given: 2figures a and b

AP=PB, AQ=QC.

To prove: PQ || BC and PQ=1/2 BC

              ▲ APQ ≅ ▲ QRC

Proof steps:

AQ=QC [midpoint]
∠ APQ = ∠QRC [Corresponding angles for parallel lines cut by an transversal].
∠PBR=∠QRC=∠APQ [Corresponding angles for parallel lines cut by an transversal].
∠RQC=∠PAQ [When 2 pairs of corresponding angles are congruent in a triangle, the third pair is also congruent.]
Therefore , ▲APQ ≅ ▲QRC
AP=QR=PB   and   PQ=BR=RC.

Since midpoints are unique, and the lines connecting points are unique, the proposition is proven.