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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
If a ray (or line) bisects an angle of a triangle, then it divides the opposite side in the ratio of the adjacent sides.
In triangle PQR, ray PM is said to bisect angle ∠QPR.
`(RM)/(MQ) = (PR)/(PQ)`
Given:
- PR = 7
- PQ = 10
- RM = 6
- MQ = 8
`(RM)/(MQ) = 6/8 = 3/4`
`(PR)/(PQ) = 7/10`
`3/4 cancel= 7/10`
So, RM : MQ ≠ PR : PQ, and hence, ray PM does not bisect ∠QPR.
Ray PM is not the bisector of ∠QPR
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In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.
Complete the proof by filling in the boxes.
In △PMQ, ray MX is bisector of ∠PMQ.
∴ `square/square = square/square` .......... (I) theorem of angle bisector.
In △PMR, ray MY is bisector of ∠PMQ.
∴ `square/square = square/square` .......... (II) theorem of angle bisector.
But `(MP)/(MQ) = (MP)/(MR)` .......... M is the midpoint QR, hence MQ = MR.
∴ `(PX)/(XQ) = (PY)/(YR)`
∴ XY || QR .......... converse of basic proportionality theorem.
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If ΔABC ∼ ΔDEF such that ∠A = 92° and ∠B = 40°, then ∠F = ?
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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]
