TRIANGLES 143
Example 9 : In Fig. 6.43, the line segment
XY is parallel to side AC of ABC and it
divides the triangle into two parts of equal
areas. Find the ratio
AX
AB
✁
Solution : We have XY || AC (Given)
So, ✂ BXY = ✂ A and✂ BYX = ✂ C (Corresponding angles)
Therefore, ABC ~ XBY (AA similarity criterion)
So,
ar (ABC)
ar (XBY) =2
AB
XB
✄ ☎
✆ ✝
✞ ✟
(Theorem 6.6) (1)Also, ar (ABC) = 2 ar (XBY) (Given)
So,
ar (ABC)
ar (XBY) =2
1
(2)
Therefore, from (1) and (2),
2
AB 2
XB 1✄ ☎
✆ ✝ ✠
✞ ✟
, i.e.,AB 2
XB 1
✡
or,
XB
AB
=
1
2
or,
1–XB
AB
=
1
1–
2
or,
AB – XB 2 1
AB 2
☞ ☛
, i.e.,AX 2 1
AB 2
☞ ☛
=
22
2
✌
.
EXERCISE 6.4
- Let ✍ ABC ~ ✍ DEF and their areas be, respectively, 64 cm^2 and 121 cm^2. If EF = 15.4
cm, find BC. - Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.
If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Fig. 6.43