Question

mid point theorem prove​

Answer 1

Answer:

Consider the triangle ABC, as shown in the above figure,

Let E and D be the midpoints of the sides AC and AB. Then the line DE is said to be parallel to the side BC, whereas the side DE is half of the side BC; i.e.

DE∥BC

DE = (1/2 * BC).

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.

Answer 2

$\huge{\underline{\underline{\mathrm{Question}}}}:-$

The line Segment joining the mid points of two side of a triangle is parallel to the third side.

$\huge{\underline{\underline{\mathrm{Solution}}}}:-$

Given :-

In ∆ABC, E and F are the mid points of sides AB and AC respectively.

To prove :-

EF || BC

To proof :-

E & F are mid points of Sides AB & AC respectively.

• AE = EB
• AF = FC

→ In ∆ AEF & ∆CDF

• BA || CD and AC is a transversal

∠EAF = ∠FAC -----(Alternate Interior Angels)

F is a mid point

∴ AF = FC

∠AFE = ∠CFD ( Pair of Opposite angles)

∆ AEF ≅ ∆ CDF -----By ASA ( Angle Side Angle) rule.

∴ EF = FD & BE = AE = CD

→ BCDE is a parallelogram

∴ED || BC

So, EF || BC

Hence Proved!

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