CHAPTER 16. GEOMETRY 16.4
Draw BE and CD.
area�ADE
area�BDE
=
1
2 AD. h^1
1
2 DB. h^1
=
AD
DB
area�ADE
area�CED
=
1
2 AE. h^2
1
2 EC. h^2
=
AE
EC
but area�BDE = area�CED (equal base and height)
∴
area�ADE
area�BDE
=
area�ADE
area�CED
∴
AD
DB
=
AE
EC
∴ DE divides AB and AC proportionally.
Similarly,
AD
AB
=
AE
AC
AB
BD
=
AC
CE
Following from Theorem1, we can prove the midpoint theorem.
Theorem 2. Midpoint Theorem: Aline joining the midpoints of two sides of a triangle is parallel to
the third side and equalto half the length of thethird side.
Proof:
This is a special case ofthe Proportionality Theorem (Theorem 1).
If AB = BD and AC = AE,
and
AD = AB + BD = 2AB
AE = AC + CB = 2AC
then DE� BC and BC = 2DE.
A
B C
D E
Theorem 3. Similarity Theorem 1:Equiangular triangles have their sides in proportion and are there-
fore similar.
A
F
H
B C
D
E
G
�
�
Given:�ABC and�DEF withAˆ =Dˆ;Bˆ =Eˆ;Cˆ =Fˆ
R.T.P.:
AB
DE
=
AC
DF
Construct: G on AB, so that AG = DE
H on AC, so that AH = DF