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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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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.
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But `(MP)/(MQ) = (MP)/(MR)` .......... M is the midpoint QR, hence MQ = MR.
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