proof of mid point theorem. iwant this answer now please
Hi Rishab!
Here is the answer to your question.
Consider a triangle ABC. Let P and Q be the mid-points of sides AB and AC respectively
In ∆APQ and ∆CRQ
∠AQP = ∠CQR (vertically opposite)
PQ = QR (by construction)
AQ = CQ (Q is the mid-point of AC)
∴ ∆APQ ≅ ∆CRQ (SAS congruence criterion)
⇒AP = CR (By C.P.C.T) … (1)
And ∠APQ = ∠CRQ (By C.P.C.T) … (2)
It is given that P is the mid-point of AB
∴ AP = PB … (3)
From (1) and (3)
PB = CR … (4)
From (2), PB || CR … (5)
From (4) and (5), PBCR is a parallelogram
∴ BC || PR and BC = PR
⇒ BC || PQ and BC = PQ + QR = PQ + PQ = 2PQ (by construction)
⇒ BC || PQ, PQ = ½ BC
Hence mid-point theorem is proved
Hope! This will help you.
Cheers!