**Mid-Point
Theorem**

According to Mid-point Theorem,

"The line segment joining the mid-points of any two
sides of a triangle is parallel to the third side and is half of it."

**Given:**D and E are mid-points of

*AB*and

*AC*respectively.

**To Prove:**(i)

*DE*‖

*BC*(ii)

*DE =*½

*BC*

**Construction:**Draw

*CF*parallel to

*BA*to meet

*DE*produced at F.

**Proof:**

1.
In

*Î”AED*and*Î”CEF,**AE = EC*(Given)

*∠*

*AED*=

*∠*

*CEF*(Vertically opp. angles)

*∠*

*DAE*=

*∠*

*FCE*(Alternate angles as

*BA*‖

*CF*and AC meets them)

Thus,

*Î”**AED*≅*Î”**CEF*(By A.S.A. congruence condition)
2.

*AD = CF*(Corresponding sides of congruent triangles)
3. But

*AD = BD*(D is mid-point of*AB*)
4.

*CF = BD*(From 2 and 3)
5. Therefore,

*DBCF*is a ‖gm. (*BD = CF*and*BD*‖*CF*)
Therefore,

*DF*‖*BC*and hence*DE*‖*BC*(Opp. sides of ‖gm are parallel)
Also

*DE = EF*(*Î”**AED*≅*Î”**CEF*)
DE = ½

*DF*= ½*BC*(Since*DF = BC*being opp. Sides of ‖gm*DFCB*)
Hence,

*DE*‖*BC*and DE = ½*BC.***Converse
of Mid-point Theorem **

The line drawn through the mid-point
of one side of a triangle, parallel to another side bisects the third side.

**Given:**

*XA = XB*and

*XY*‖

*BC*of triangle

*ABC*.

**To Prove:**AY = YC.

**Construction:**Draw

*CZ*parallel to

*BA*to meet

*XY*produced at Z.

**Proof:**

*1.*

*XBCZ*is a parallelogram (

*XZ*‖

*BC*(given) and BX ‖ CZ)

*2.*

*BX = CZ*(Opposite sides of ‖gm XBCZ)

*3.*

*XA = CZ*(XA = BX, given)

*4.*In

*Î”AYX*and

*Î”CYZ,*

*XA = CZ*(From statement (3))

*∠*

*XAY*=

*∠*

*ZCY*(Alternate angles as

*BA*‖

*CZ*and AC meets them)

*∠*

*AXY*=

*∠*

*CZY*(Alternate angles as

*BA*‖

*CZ*and

*XZ*meets them)

Thus,

*Î”**AYX**≅**Î”**CYZ*(By A.S.A.)
Hence,

*AY = YC*(CPCT)**Intercept
Theorem**

According to Intercept Theorem,

"If
a transversal makes equal intercepts on three or more parallel lines, then any
other line cutting them will also make equal intercepts."

**Given:**

*AB*‖

*CD*‖

*EF*. The transversal

*L*cuts them in such a way that intercept

_{1}L_{2}*AC*= intercept

*CE*.

**To Prove:**

*BD = DF*where

*M*is another transversal meeting them at B, D, F, respectively.

_{1}M_{2}**Construction:**Through A and C, draw

*AG*and

*CH*parallel to the line

*BDF*to cut

*CD*at G and

*EF*at H.

**Proof:**

1.
In

*Î”ACG*and*Î”CEH,**AC = CE*(Given)

*∠*

*ACG*=

*∠*

*CEH*(Corr. angles as

*CD*‖

*EF*and

*ACE*meets them)

*∠*

*CAG*=

*∠*

*ECH*(Corr. angles as

*AG*‖

*BF*‖

*CH*and

*ACE*meets them)

Thus,

*Î”**ACG**≅**Î”**CEH*(By*A.S.A.)*
2.

*Therefore,**AG = CH*(By CPCT)
3.

*AGDB*is a ‖gm (Both pair of opp. sides are parallel)
Thus,

*AG = BD*(Opp. sides of ‖gm are equal)
4.

*CHFD*is a ‖gm (Both pairs of opp. sides are parallel)
Therefore,

*CH = DF*(Opp. sides of ‖gm are equal)
Hence, from (2), (3) and (4), it
follows that

*BD = DF*.

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