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प्रश्न
Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR.
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उत्तर
In ∆ PQR,
PQ = 7, PR = 3, QM = 3.5, and MR = 1.5 ...(Given)
`"PQ"/"PR" = 7/3` ...(i)
`"QM"/"MR" = 3.5/1.5 = (3.5 × 10)/(1.5 × 10) = 35/15 = 7/3` ...(ii)
From (i) and (ii)
∴ `"PQ"/"PR" = "QM"/"MR"`
∴ by converse of angle bisector theorem,
∴ Ray PM is the bisector of ∠QPR.
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If ΔABC ∼ ΔDEF such that ∠A = 92° and ∠B = 40°, then ∠F = ?
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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]
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)]
