(Q16) In trapezium ABCD, AB ∥ DC. E and F are points on non-parallel sides AD and BC respectively such that EF ∥ AB. Show that

AE

ED

=

BF

FC

Solution :

Let us join A, C to intersect EF at G

AB ∥ DC and EF ∥ AB (given)

⇒ EF ∥ DC (Lines parallel to the same line are parallel to each other)

In ∆ADC, EG ∥ DC

So

ED

=

AG

(by BPT)   ....(1)

Similarly, In ∆CAB, GF ∥ AB

CG

=

FB

(by BPT) i.e.,

AG

GC

=

BF

FC

 ....(2)
From (1) & (2)

AE

=

FC