Maths

TRIANGLES 127
Example 2 : ABCD is a trapezium with AB || DC.
E and F are points on non-parallel sides AD and BC
respectively such that EF is parallel to AB
(see Fig. 6.14). Show that AE BF
ED FC .
Solution : Let us join AC to intersect EF at G
(see Fig. 6.15).
AB || DC and EF || AB
(Given)
So, EF || DC (Lines parallel to the same line are
parallel to each other)
Now, in ✁ ADC,
So,
EG || DC
AE
ED = AG
GC
Similarly, from ✁ CAB,
(As EF || DC)
(Theorem 6.1) (1)
CG
AG = CF
BF
Fig. 6.14
Fig. 6.15
i.e.,
Therefore, from (1) and (2),
AG
GC = BF
FC
AE
ED = BF
FC
(2)
Example 3 : In Fig. 6.16, PS
SQ = PT
TR
and ✂ PST =
✂ PRQ. Prove that PQR is an isosceles triangle.
Solution : It is given that PS
SQ
✄
PT
TR
So, ST || QR (Theorem 6.2)
☎
Fig. 6.16
Therefore, ✂ PST = ✂ PQR (Corresponding angles) (1)