Question

state and verify the mid point therom​

Answer 1

Mid point theorem :

The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.

To prove :

1. DE || BC
2. DE = ½ CB

Construction :

Extend DE upto F so that DE = EF. Join F to C.

Proof :

In △ AED and △ CEF

⇒ ∠AED = ∠CEF (Vertically Opposite Angle)

⇒ DE = EF (By construction)

⇒ AE = EC (E is the mid point of AC)

∴ △AED ≅ △CEF (By S.A.S)

Hence,

∠1 = ∠2 (By Alternate Interior Angle)

∴ AD || FC

⇒ BD || FC ....i)

Again,

⇒ AD = FC (By c.p.c.t)

⇒ BD = FC ....ii) (AD = BD)

From equation i) and ii),

⇒ BDFC is a paralellogram

∴ DF || BC (Property of paralellogram)

∴ DE || BC ...... a)

And,

⇒ DE = BC (Property of paralellogram)

⇒ DE + EF = BC

⇒ DE + DE = BC (DE = EF)

⇒ 2DE = BC

⇒ DE = BC/2 ....b)

∴ It is proved from equation a) and b) that,

• DE || BC &
• DE = BC/2

Answer 2

Given

ABCD is a triangle where E and F are mid point

AB and 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 canstruction)

With transversal ED.

Angle AEF= Angle CDE ( Alternate angle)---(1)

In ∆ AEF and ∆ CDE ( from (1))

Angle AEF= Angle CED.(vertically opposite angle

AF =CE ( As F is mid point of AC)

•°• ∆ ≈ ∆ CDE (AAS rule)

So, EA= DC (CPCT)

But ,EA=EB

Hence EB=DC. (E is mid point of AB)

Now

In EDCD

EB || DC & EB = DC.

Thus one pair of opposite sides is equal and

parallel Hence EBCD is a parallelogram.

Since opposite side of parallelgram are parallel.

So, ED || BC

EF || BC

Hence proved ✓✓✓✓✓✓✓