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प्रश्न
In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.
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उत्तर
In △ABC,
seg BD bisects ∠ABC. ...(Given)
∴ by the theorem of angle bisector of a triangle,
∴ `"AB"/"BC" = "AD"/"DC"`
∴ `x/(x + 5) = (x – 2)/(x + 2)`
∴ x(x + 2)= (x – 2)(x + 5)
∴ x2 + 2x = x(x + 5) - 2(x + 5)
∴ x2 + 2x = x2 + 5x - 2x - 10
∴ x2 + 2x = x2 + 3x - 10
∴ x2 - x2 + 2x - 3x = - 10
∴ - x = - 10
∴ x = 10
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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.
∴ `("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]
