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give the proof of the mid-point theorum.

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 equal to half of it".

⇒ ED | | BC  (Opposite sides of parallelogram are parallel)

⇒ EF | | BC 

EF = DF  (Proved)

⇒ EF + DF = ED = BC  (Opposite sides of the parallelogram are equal)

⇒ EF + EF = BC

⇒ 2 EF = BC 

 

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 here you gooo!!!!!!!,..........

                     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
 
 
                   hhhe 
⇒ BC || PQ and BC = PQ + QR = PQ + PQ = 2PQ (by construction)
⇒ BC || PQ, PQ = ½ BC
Hence mid-point theorem is proved
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