Question

prove that-the line segment joining the midpoints of two sides of a triangle is equal to half the third side of triangle​

Answer 1

Mid Point Theorem

✧ Mid Point Theorem: Mid point theorem says that “The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half the third side.

✧ Given: Let ABC is a triangle, E & F are the mid points of the side AB and AC of the triangle.

✧To Prove:  $\sf{EF \parallel BC\;and\;EF=\dfrac{1}{2} BC}$

✧ Construction:  Draw a line CD parallel to AB and it intersects the extended EF at D.

✧ Proof:

In ΔAEF and ΔCDF,

⇒ ∠EAF =∠FCD                       [Alternate interior angles]

⇒ AF = FC                                [F is mid point of AC]

⇒ ∠AFE = ∠CFD                      [Vertically opposite angle]

∴ ΔAEF ≅ ΔCDF                      [ASA concurency rule]

So, EF = DF and AE = DC       [By CPCT]

∴ BE = AE = DC

Therefore BCDE is a parallelogram.

⇒ ED║BC              [Opposite sides of parallelogram are ║]

⇒ EF║BC

∴ ED = ED

Now, EF + DF = DE = BC     [Opposite sides of parallelogram are equal]

⇒ EF + EF = BC                    [DF = EF]

⇒ 2EF = BC

$\sf{\implies EF = \dfrac{1}{2}BC}$

Hence Proved!!!

Answer 2

☆Question:-

prove that-the line segment joining the midpoints of two sides of a triangle is equal to half the third side of triangle

☆Solution:-

▪Given :-

• ABCD is a triangle where E and F are mid points of AB and AC respectively.

▪To prove :-

• $EF || BC$

▪Construction :-

• Through C draw a line segment parallel to AB and extend EF to meet this line at D.

▪Proof :-

• Since AB is parallel to CD. ( By construction )

with transversal ED.

• Angle AEF = angle CDF ( alternate angles.)

In triangle AEF and CDF :-

• Angle AEF = CDF

• Angle AFE = CFD

• AF = CF

$\therefore$ Triangle AEF is not congruent to CDF. (AAS rule)

So,

EA = DC but EA = EB.

Hence,

EB = DC.

In EBCD, EB is parallel to DC and EB = DC

One pair of opposite sides is equal and parallel.

Hence, EBCD is a parallelogram.

Since, opposite sides of parallelogram are parallel.

So, ED || BC.

i.e EF || BC.

Hence, proved