Question

State and prove mid point theoram

Answer 1

Mid point theorem states that a line joining the mid points of two sides of a triangle is always parallel to the third side and is half of the third side.

For proof, refer to the attachment

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Answer 2

$\bigstar\rm\blue{GIVEN}$

• In Fig 1

• AB=AC ,AD=AE....1

• DE||BC.

$\bigstar\rm\blue{TO\:PROVE}$

• DE||BC

• DE=1/2BC

$\bigstar\rm\red{Construction}$

• Expand DE to F and Meet F to C.(see Fig.2)

• CF||AB OR CF||DB

Now,

In ∆sADE&ECF.

• $\rm{\angle{AED}=\angle{FEC}}$(Vertically.Opps.angle)

• $\rm{\angle{EAD}=\angle{ECF}}$(Alt.interor angle).

• $\rm{AD=CE}$...From1

So, ∆ADE IS CONGURENT TO ∆ECF BY ASA.

From this we get,

• AD= CF or DB=CF

So,

BDFC is Parallelogram.

in which DF||BC and DF=BC

we can also Write DF as DE as E is the mid point so, First one is proved

NOW,

DF=BC

DE+FE=BC

2DE=BC

DE=1/2BC

Hence, Proved.