untitled

Hence, the area of the triangle ABC = auh
2

1

and the area of the triangle

DEF= duh
2

1

.

Therefore, area of the triangle ABC : area of the triangleDEF = auh
2

1

: duh

2

1

=a:d=BC:EF that is, the areas and bases ate proportional.
(2) If the bases of two triangles are equal, their heights and areas are
proportional.

Let the heights of the triangles ABC and DEF be AP =h,DQ=krespectively and

the base in both cases be b. Hence, the area of the triangle ABC = buh
2

1

and the

area of the triangle DEF= buk
2

1

Therefore, area of the triangle ABC : area of the triangleDEF = buh
2

1

: buk

2

1

=h: k = AP: DQ
Theorem 1
A straight line drawn parallel to one side of a triangle intersects the other two
sides or those sides produced proportionally.
Proposition : In the figure, the straight
line DE is parallel to the side BCof the
triangle ABC.DE intersects AB and AC
or their produced sections at DandE
respectively. It is required to prove that,
AD : DB = AE : EC.
Construction: Join B, E and C, D.
Proof:
Steps Justificaltin
(1) The heights of 'ADE and'BDE are
equal.

[The bases of the triangles of equal height

are proportional]

Fig-1 Fig-2