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Mid-Point Theorem
STATEMENT: It states that a line segment drawn from the midpoints of any two sides of a triangle is
parallel to the third side of the triangle and also it is half the length of the third side.
A
D E M
B C
Given: In ABC, D is the midpoint of AB and E is the midpoint of AC
So, AD = DB & AE= EC
To Prove: DE// BC & BC = 2DE
Construction : Let DE is extended to a point M such that DE = EM
Proof:
In CME & ADE
CE = AE (Given
AED = CEM (Vertically Opposite Angles)
ME = DE (Construction)
So, By Side-Angle-Side (SAS) Similar Property ABC & ADE are Congruent
ABC ADE
Therefore, CME = ADE , But these are alternate interior angles
So, DM // BC means DE // BC.
Which means BDMC is a parallelogram.
Now, In parallelogram BDMC,
DM = BC (Parallel & Opposite sides are equal)
Also, DM = DE + EM
DM = DE + DE (DE= EM )
DM = 2DE
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Hence BC= 2DE
That is DE = ½ BC
HENCE PROOVED

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Mid-Point Theorem STATEMENT: It states that a line segment drawn from the midpoints of any two sides of a triangle is parallel to the third side of the triangle and also it is half the length of the third side. A D E M B C Given: In ∆ABC, D is the midpoint of AB and E is the midpoint of AC So, AD = DB & AE= EC To Prove: DE// BC & BC = 2DE Construction : Let DE is extended to a point M such that DE = EM Proof: In ∆CME & ∆ADE CE = AE (Given ∠AED = ∠CEM (Vertically Opposite Angles) ME = DE (Construction) So, By Side-Angle-Side (SAS) Similar Property ABC & ADE are Congruent ∆ABC ≅ ∆ADE Therefore, CME = ADE , But these are alternate interior angles So, DM // BC means DE // BC. Which means BDMC is a parallelogram. Now, In parallelogram BDMC, DM = BC (Parallel & Opposite sides are equal) Also, DM = DE + EM DM = DE + DE DM = 2DE (DE= EM ) Hence BC= 2DE That is DE = ½ BC HENCE PROOVED Name: Description: ...
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