Question

state and prove midpoint theorem

Answer 1

The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

Given: In triangle ABC, P and Q are mid-points of AB and AC respectively.

To Prove: i) PQ || BC ii) PQ = 1 2 BC

Construction: Draw CR || BA to meet PQ produced at R.

Proof:
∠QAP = ∠QCR (Pair of alternate angles) ---------- (1)

AQ = QC (∵ Q is the mid-point of side AC) ---------- (2)

∠AQP = ∠CQR (Vertically opposite angles) ---------- (3)

Thus, ΔAPQ ≅ ΔCRQ (ASA Congruence rule)

PQ = QR (by CPCT) or PQ = 1 2 PR ---------- (4)

⇒ AP = CR (by CPCT) ........(5)

But, AP = BP (∵ P is the mid-point of the side AB)

⇒ BP = CR

Also. BP || CR (by construction)

In quadrilateral BCRP, BP = CR and BP || CR

Therefore, quadrilateral BCRP is a parallelogram.

BC || PR or, BC || PQ

Also, PR = BC (∵ BCRP is a parallelogram)

⇒ 1 2 PR = 1 2 BC

⇒ PQ = 1 2 BC [from (4)]

Answer 2

Hey !

______________

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.

# Hope it helps #