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Using the basic proportionality theorem, prove that a line drawn through the midpoints of one side of a triangle is parallel to the other side that bisects the third side.
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Here is your solution
From the given diagram we get,
The triangle PQR in which S is the midpoint of P and Q such that PS=SQ
A line parallel to QR intersects PR at T such that ST ǁ QR
To prove: T is the midpoint of PR
Proof: S is the midpoint of PQ
Therefore, PS=SQ
=>PS/QS = 1 – – – – – – – (1)
In triangle PQR, ST ǁ QR,
Therefore, PS/SQ = PT / TR [Using the basic proportionality theorem]
=>1 = PT/ TR [from equation (1)]
Therefore, PT = TR
Hence, T is the midpoint of PR
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Question:-
Using the basic proportionality theorem, prove that a line drawn through the midpoints of one side of a triangle is parallel to the other side that bisects the third side.
Answer:-
Given,ΔABC in which D is the mid point of AB such that,AD=DB
A line parallel to BC intersect AC At E as shown in above figure such that,DE||BC.
To prove E is the mid point of AC
Proof: D is the mid point of AB
Therefore, AD=DB
AD/DB=1____[1]
In Δ ABC,DE||BC
Therefore,AD/DB=AE/EC [by using basic proposnality theorm]
1=AE/EC (From eq 1st)
Therefore,AE=EC
Hence,E is the mid point of AC.
Therefore,proved.
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