Question

the line joining midpoint of the two sides of a triangle is parallel to the third side and is of it is

Answer 1

Answer:

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Step-by-step explanation:

Given: ABCD is a Triangle where E & F are Mid points of AB & AC respectively

To Prove: EF║BC

Construction: Through C draw a Line Segment Parallel to AB & Extend EF To meet This Line at D

Proof: Since AB║CD  ( By Construction )

          With Transversal AD

∠AEF = ∠CDF       (Alternate Angles)......( 1 )

In Δ AEF & ΔCDF

In ∠ AEF & ∠CDF .....From ( 1 )

∠ AFE & ∠CFD   ......  ( Vertically opposite Angle )

AF = CF   ( As F is mid point of AC )

Δ AFE ≅ ΔCFD    ( AAS Rule )

So, EA = DC    (CPCT)

But, EA = EB   ( E is the mid point of AB )

Hence, EB = DC

Now,

In EBCD,

EB ║DC & EB = DC

Thus, 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

Answer 2

heya meta...here is the solution ✌️✌️✌️




Given: ABCD is a Triangle where E & F are Mid points of AB & AC respectively

To Prove: EF║BC

Construction: Through C draw a Line Segment Parallel to AB & Extend EF To meet This Line at D

Proof: Since AB║CD  ( By Construction )

          With Transversal AD

∠AEF = ∠CDF       (Alternate Angles)......( 1 )

In Δ AEF & ΔCDF

In ∠ AEF & ∠CDF .....From ( 1 )

∠ AFE & ∠CFD   ......  ( Vertically opposite Angle )

AF = CF   ( As F is mid point of AC )

Δ AFE ≅ ΔCFD    ( AAS Rule )

So, EA = DC    (CPCT)

But, EA = EB   ( E is the mid point of AB )

Hence, EB = DC


Now,

In EBCD,

EB ║DC & EB = DC

Thus, 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



HOPE you understand ❤️❤️❤️