Question

State and proove mid point theorem

Answer 1

$\large{\texttt{Midpoint Theorem}}}$

The linesegment joining midpoints of two sides of a triangle is parallel to the third side and half of the third side.

$\large{\large{\texttt{Proof}}}$

Given

In ΔABC,

AD = DB

AE = AC

To Prove

DE║BC

DE = $\sf{\frac{1}{2}}$ BC

Construction

Draw a line through C, parallel to BA

Proof

Consider ΔADE & ΔCFE

AE = CE  [Given]

∠1 = ∠2   [Alternate interior angles]

∠3 = ∠4  [V.O.A]

$\implies$ ΔADE ≅ ΔCFE by ASA congruency.

AD = CF [CPCT] → 1

AD = DB [Given] → 2

From 1 and 2 we can say that,

DB = CF [Things which are equal to the same thing are equal]

DB ║CF [Construction]

∴ DBCF is a Parallelogram

[Since 1 pair of opposite sides are equal and parallel]

$\implies$ DF = BC

[Opp. sides of a Parallelogram]

ΔADE ≅ ΔCFE [Proved]

DE = FE [CPCT]

DE = $\sf{\frac{1}{2}$ DF [DE = FE]

But, DF = BC

$\large{\boxed{\sf{DE = \frac{1}{2} \ BC}}$

DF║BC

[Opposite sides of a Parallelogram are parallel]

But, DF = DE

$\implies$ DE║BC

 

Hence, Proved.

Answer 2

MidPoint Theorem Statement

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

Construction- Extend the line segment DE and produce it to F such that, EF=DE.

In the triangle, ADE, and also the triangle CFE

EC= AE —– (given)

∠CEF = ∠AED {vertically opposite angles}

EF = DE { by construction}

hence,

△ CFE ≅ △ ADE {by SAS}

Therefore,

∠CFE = ∠ADE {by c.p.c.t.}

∠FCE= ∠DAE {by c.p.c.t.}

and CF = AD {by c.p.c.t.}

The angles, ∠CFE and ∠ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.

In a similar way, ∠FCE and ∠DAE are the alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.

Therefore, CF ∥ AB

So, CF ∥ BD

and CF = BD {since BD = AD, it is proved that CF = AD}

Thus, BDFC forms a parallelogram.

By the use of properties of a parallelogram, we can write

BC ∥ DF

and BC = DF

BC ∥ DE

and DE = (1/2 * BC).

Hence, the midpoint theorem is Proved.