state and prove midpoint theorem by asa rule
Answers
Step-by-step explanation:
The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
Anytime you have a line segment that connects two sides of a triangle at the midpoints, you automatically know that the sides are cut in half, and that the segment is parallel to the third side of the triangle. Parallel sides are shown by using this symbol ||. You also know the line segment is one-half the length of the third side.
Midpoint Theorem
Midpoint theorem congruent sides
This indicates that points R and S are midpoints of sides AT and AV, respectively. From the Midpoint Theorem, since the segment RS connects the two sides at the midpoints, then RS || TV and RS is one-half the length of side TV.
This theorem allows us to prove some things about the triangle. First, if we know the length of TV, then we can figure out the length of RS, and vice-versa, since RS = ½(TV). It also allows us to find the lengths of AS, VS, TR and AR. Since RS is parallel to TV, then we also know the distance between these two line segments are equal.
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Answer:
Step-by-step explanation:
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..
Thanking you.
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Aryan Bhardwaj
Aryan Bhardwaj, Student
Answered Aug 27, 2018
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)]
!!
t Deen Dayal Upadhyay Gorakhpur University
Answered Aug 24, 2018
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ArPit CheChani
ArPit CheChani, Bachelor of technology Mechanical Engineering & Mathematics, Swami Keshvanand Institute of Technology, Mana...
Answered Dec 20, 2018
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
1. 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]
4. So,Triangle AEF = CDF(Angle side Angle rule)
5. EF=FD AND AE = CD (c.p.c.t)
6. AE=BE(E is midpoint of AB)
7. 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)..
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Aakash Singh
Aakash Singh
Answered Aug 22, 2018
AM=BM [midpoint].
< AMP = <MBQ [Corresponding angles for parallel lines cut by an transversal].
<BQM=<QCP=<APM [Corresponding angles for parallel lines cut by an transversal].
<BMQ=<MAP [When 2 pairs of corresponding angles are congruent in a triangle, the third pair is also congruent.]
AMP is congruent to MBQ. [ASA]
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Mano Kumar
Mano Kumar
Answered Aug 25, 2018
(X1+X2/2), (Y1+Y2/2)
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