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Question : 13 of 160
Marks:
+1,
-0
Solution:
(c) It is given that for a quadrilateral
ABCD,
A B=a, B C=b, A D=b-a
= ∴ According to the question,
BM=‌‌ and ‌‌=‌‌‌{∵DN=(‌)DM given
} ∴‌‌AN−AD‌‌=‌(AM−AN) ⇒‌‌5AN‌‌=4AM+AD=4(a+‌)+(b−a) ‌‌=4a+2b+b−a=3(a+b)=3AC
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