The Mid Point Theorem states that, “*The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.”*

If we consider a Δ ABC with P and Q as mid points of AB and BC respectively as shown, we are given that

AP = PB (P is the mid point)

AQ = QC (Q is the mid point)

In order to verify the statement, we have to prove that

PQ || BC

PQ= ^{1}⁄_{2} BC

Construction is very important in this particular proof. So, we have to extend PQ such that PQ = QR. Later, we have to join RC too.

In Δ ADE and CEF, we can observe that

∠1 = ∠2 (Vertically opposite ∠s)

AQ = QC (Given)

PQ = QR (Construction)

Therefore, Δ ADE ≅ ΔCEF (SAS≅), AP = CR and ∠3 = ∠4 (c.p.c.t.)

Now, we know that P is the mid point of AB

AP = PB

PB = CR (because AP = CR)

∠3 = ∠4 (proven)

We can see that ∠3 and ∠4 are alternate interior angles for lines AB and CR

PB || CF and PB = CF

We can also say that

PR || BC and PR = ^{1}⁄_{2} BC

Therefore, **PQ || BC.**

We have proven half of the theorem, now we have to show PQ= ^{1}⁄_{2} BC

PR = PQ + QR but PQ = QR

So, PR = 2 PQ

BC = 2 PQ

Therefore, **PQ = **^{1}⁄_{2} BC.

That’s it! I hope you understood the theorem. 🙂

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