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Question
Measures of some angles in the figure are given. Prove that `"AP"/"PB" = "AQ"/"QC"`.
Solution
Given: ∠APQ = 60∘, ∠ABC = 60∘
To Prove: `"AP"/"PB" = "AQ"/"QC"`.
Proof:
∠APQ = ∠ABC = 60∘ ...(Given)
∴ ∠APQ ≅ ∠ABC
∴ Seg PQ || Seg BC ...(Corresponding angles test for parallel lines )(I)
In ΔABC,
Seg PQ || Seg BC ...[From I]
By Basic proportionality theorem,
∴ `"AP"/"PB" = "AQ"/"QC"`
Hence proved.
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∴ `square/square = square/square` .......... (II) theorem of angle bisector.
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In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find `"AX"/"XY"`.
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In Δ ABC and Δ PQR,
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In ΔABC, ray BD bisects ∠ABC.
If A – D – C, A – E – B and seg ED || side BC, then prove that:
`("AB")/("BC") = ("AE")/("EB")`
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In ΔABC, ray BD bisects ∠ABC.
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∴ `("AE")/("EB") = ("AD")/("DC")` ....(ii) `square`
∴ `("AB")/square = square/("EB")` [from (i) and (ii)]
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From the information given in the figure, determine whether MP is the bisector of ∠KMN.
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Complete the proof by filling in the boxes.
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]
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)]
