proof of mid point theorem
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Mid-Point Theorem :-
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)]
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)]
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