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Question
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.
Solution
In ΔABC, ray BD is the bisector of ∠ABC
∴ by the theorem of an angle bisector of a triangle,
`"AB"/"BC"="AD"/"DC"` ...(1)
In ΔABC, ray CE is the bisector of ∠ACB
∴ by the theorem of an angle bisector of a triangle,
`"AC"/"BC"="AE"/"EB"` ...(2)
Seg AB ≅ seg AC (Given) ...(3)
[From (1), (2) and (3)]
∴ `"AB"/"BC"="AC"/"BC"` ...(4)
In ΔABC,
`"AE"/"EB"="AD"/"DC"` ...[From (1), (2) and (4)]
∴ by converse of basic proportionality theorem,
seg ED || side BC
∴ ED II BC
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