State and prove mid point theorem
Answers
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)]
You may study this from the link below:
Thank you!!
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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)]
You may study this from the link below:
Thanking you.
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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.
Take a triangle ABC,E and F are the mid-points of side AB and AC resp.
Construction:-Through C,draw a line II BA to meet EF produced at D.
Proof:-
In Triangle AEF and CDF
AF=CF(F is midpoint of AC)
2. Angle AFE= Angle CFD (Vertically opp. angles)
3. Angle EAF= Angle DCF [Alt. angles,BA II CD(by construction) and AC is a transversal]
So,Triangle AEF = CDF(Angle side Angle rule)
5. EF=FD AND AE = CD (c.p.c.t)
AE=BE(E is midpoint of AB)
BE=CD(from 5 and 6)
8.EBCD is a IIgm [BA II CD (by construction) and BE = CD(from 7)]
9.EF II BC AND ED=BC (Since EBCD is a IIgm)
10.EF = 1/2 ED (Since EF = FD,from 5)
11.EF = 1/2 BC (Since ED = BC,from 9)
Hence,EF II BC AND EF = 1/2 BC which proves the mid-point theorem.
(ans)..
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.
Here, In △ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined.
Given: AD = DB and AE = EC.
To Prove: DE ∥ BC and DE = 12 BC.
Construction: Extend line segment DE to F such that DE = EF.
Proof: In △ ADE and △ CFE
AE = EC (given)
∠ AED = ∠ CEF (vertically opposite angles)
DE = EF (construction)
hence
△ ADE ≅ △ CFE (by SAS)
Therefore,
∠ ADE = ∠ CFE (by c.p.c.t.)
∠ DAE = ∠ FCE (by c.p.c.t.)
and AD = CF (by c.p.c.t.)
The angles ∠ ADE and ∠ CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF.
Similarly, ∠ DAE and ∠ FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC.
Therefore, AB ∥ CF
So, BD ∥ CF
and BD = CF (since AD = BD and it is proved above that AD = CF)
Thus, BDFC is a parallelogram.
By the properties of parallelogram, we have
DF ∥ BC
and DF = BC
DE ∥ BC
and DE = 12 BC (DE = EF by construction)
l have attched a figure.
hope it helps you
