prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side
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We have to prove that line segment joining the mid points of two sides of triangle is equal to the half of the third side of the triangle.
Let ABC be any triangle. Also, let D and E be the midpoints of AB and AC respectively.
DE is expanded to F such that EF = DE
In ∆ADE and ∆CEF,
AE = EC
DE = EF and
∠AED = ∠CEF (vertically opposite angles)
Therefore, by SAS criterion ∆ADE ≅ ∆CEF
∴ CF = AD (By CPCT)
⇒ BD = CF (1)
Also ∠ADE = ∠EFC (By CPCT)
But ∠ADE and ∠EFC are alternative interior angles on AD and CF.
⇒ AD || CF
⇒ BD || CF (2)
When two sides of a quadrilateral are parallel and equal, it is a parallelogram.
From (1) and (2) it is proved that DFCB is a parallelogram
∴ DF = BC and DF || BC
Also, 2 DE = DF = BC
⇒ DE =
Let ABC be any triangle. Also, let D and E be the midpoints of AB and AC respectively.
DE is expanded to F such that EF = DE
In ∆ADE and ∆CEF,
AE = EC
DE = EF and
∠AED = ∠CEF (vertically opposite angles)
Therefore, by SAS criterion ∆ADE ≅ ∆CEF
∴ CF = AD (By CPCT)
⇒ BD = CF (1)
Also ∠ADE = ∠EFC (By CPCT)
But ∠ADE and ∠EFC are alternative interior angles on AD and CF.
⇒ AD || CF
⇒ BD || CF (2)
When two sides of a quadrilateral are parallel and equal, it is a parallelogram.
From (1) and (2) it is proved that DFCB is a parallelogram
∴ DF = BC and DF || BC
Also, 2 DE = DF = BC
⇒ DE =
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Step-by-step explanation:
I explained this with using the coordinate geometry with slope formula.
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