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Question
In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:
Proof :
In ΔABC, ray BD bisects ∠B.
∴ `square/("BC") = ("AD")/("DC")` ...(I) (`square`)
ΔABC, DE || BC
∴ `(square)/("EB") = ("AD")/("DC")` ...(II) (`square`)
∴ `("AB")/square = square/("EB")` ...[from (I) and (II)]
Solution
In ΔABC, ray BD bisects ∠B.
∴ `(bbunderline"AB")/("BC") = ("AD")/("DC")` ...(I) (By angle bisector theorem)
In ΔABC, DE || BC
∴ `(bbunderline"AE")/("EB") = ("AD")/("DC")` ...(II) (Basic proportionality theorem)
∴ `("AB")/bbunderline"BC" = bbunderline"AE"/("EB")` ...[from (I) and (II)]
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solution:
In ∆PMQ,
Ray MX is the bisector of ∠PMQ.
∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]
Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.
∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]
But `("MP")/("MQ") = ("MP")/("MR")` .............(III) [As M is the midpoint of QR.]
Hence MQ = MR
∴ `("PX")/square = square/("YR")` .............[From (I), (II) and (III)]
∴ XY || QR .............[Converse of basic proportionality theorem]
