Mid-Point Theorem, Intercept Theorem

# Mid-Point Theorem, Intercept Theorem

## 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 ΔAED and ΔCEF,

AE = EC                          (Given)

AED = CEF                  (Vertically opp. angles)

DAE = FCE                 (Alternate angles as BA CF and AC meets them)

Thus, ΔAED ΔCEF       (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                    (ΔAED ΔCEF)

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 ΔAYX and ΔCYZ,
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, ΔAYX ΔCYZ                   (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 L1L2 cuts them in such a way that intercept AC = intercept CE.

To Prove: BD = DF where M1M2 is another transversal meeting them at B, D, F, respectively.

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 ΔACG and ΔCEH,

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, ΔACG ΔCEH                (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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