Question

How to prove mid point theorem​

Answer 1

Given ABC is a triangle E and F are midpoints of the sides AB ,AC respectively.

To prove : EF|| BC and EF = ½ BC. Construction : Draw a line CD parallel to AB ,it intersects EF at D. Proof : In a ΔAEF and ΔCDF ∠EAF = ∠FCD ( Alternative interior angles) AF = FC ( F is the midpoint) ∠AFE = ∠CFD ( vertically opp. Angles) ΔAEF ≅ ΔCDF (ASA congruence property) So that EF = DF and AE = CD ( By CPCT ) BE = AE = CD ∴ BCDE is parallelogram. ⇒ ED | | BC (Opposite sides of parallelogram are parallel) ⇒ EF | | BC ∴ EF = DF (Proved) ⇒ EF + DF = ED = BC (Opposite sides of the parallelogram are equal) ⇒ EF + EF = BC ⇒ 2 EF = BC ∴ EF = (1/2) BC.

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Answer 2

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.