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!

  • 11

 WHICH THEORM 1 OR 2

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 go away and ask to the expert

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to prove mid-point theorem, it is required to prove PQ.BC and PQ=BC

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how can we prove PQ = BC???

it shud be PQ = 1/2 BC... ryt???

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