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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