Assertion (A): ABCD is a trapezium with DC || AB E and F are point on AD and BC respectively, such that

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Solution

Q.
Assertion (A): ABCD is a trapezium with DC || AB. E and F are point on AD and BC respectively, such that EF || AB. Then $\frac{A E}{E D} = \frac{B F}{F C}$ .

Reason (R): Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally

1 Both A and R are true, and R is the correct explanation of A.

2 Both A and R are true, but R is not the correct explanation of A.

3 A is true, but R is false.

4 A is false, but R is true

Ans.
(1)

We joined AC which intersects EF at O.

$A s, E F ∥ A B \Rightarrow E F ∥ D C (∵ A B ∥ D C)$

$I n Δ A D C, w e h a v e E O ∥ D C$

$\frac{A E}{E D} = \frac{A O}{O C} (i)$

$A l s o, i n Δ A B C, w e h a v e A B ∥ O F$

$\frac{A O}{O C} = \frac{B F}{F C} (i i) F r o m (i) a n d (i i), w e g e t :$

$\frac{A E}{E D} = \frac{B F}{F C}$

$T h u s, A s s e r t i o n (A) i s t r u e a n d R e a s o n (R) i s a l s o t r u e, a n d R e a s o n (R) i s t h e c o r r e c t e x p l a n a t i o n o f A s s e r t i o n (A) . T h e r e f o r e, o p t i o n (a) i s c o r r e c t .$

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