Question

Prove the converse of 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 

Answer 2

Given: In ∆PQR, S is the midpoint of PQ, and ST is drawn parallel to QR.

Converse of Midpoint Theorem Proof

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To prove: ST bisects PR, i.e., PT = TR.

Construction: Join SU where U is the midpoint of PR.

Converse of Midpoint Theorem

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Proof:

Statement

Reason

1. SU ∥ QR and SU = 12QR.

1. By Midpoint Theorem.

2. ST ∥QR and SU ∥ QR.

2. Given and statement 1.

3. ST ∥ SU.

3. Two lines parallel to the same line are parallel themselves.

4. ST and SU are not the same line.

4. From statement 3.

5. T and U are coincident points.

5. From statement 4.

6. T is the midpoint of PR (Proved).

6. From statement 5.