If AD is external bisector of ∠A which meets BC at D and CE || DA. Then, which of the following is true?
A
BDCD=ABAC
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B
BDCE=ABAD
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C
Both (a) and (b)
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D
None of these
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Solution
The correct option is ABDCD=ABAC
Given, AD bisects ∠A externally and CE || AD
⇒2=∠4and∠3=∠1 [BK, CA respectively being the transversal] But ∠1=∠2 [Since AD bisects ∠A externally] ∴∠3=∠4 In ΔACE, ∠3=∠4⇒AE=AC ...(i) [∵ sides opposite to equal angles in a triangle are equal] Now, in ΔBAD, EC||AD ⇒BDCD=BAEA [by basic proportionally theorem] ⇒BDCD=ABAC [from Eq. (i)] Hence option A is the correct option.