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प्रश्न
In the figure, ray YM is the bisector of ∠XYZ, where seg XY ≅ seg YZ, find the relation between XM and MZ.
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उत्तर १
Given : YM bisects ∠XYZ, XY = YZ
In ΔXYZ,
`(XY)/(YZ)=(XM)/(MZ)` (Angle bisector theorem)
1 = `(XM)/(MZ) ` (XY = YZ)
XM = MZ
उत्तर २
Given : YM bisects ∠XYZ, XY = YZ
In ΔXYZ,
`(XY)/(YZ)=(XM)/(MZ)` (Angle bisector theorem)
1 = `(XM)/(MZ) ` (XY = YZ)
XM = MZ
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solution:
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∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]
Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.
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But `("MP")/("MQ") = ("MP")/("MR")` .............(III) [As M is the midpoint of QR.]
Hence MQ = MR
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∴ XY || QR .............[Converse of basic proportionality theorem]
