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In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x. - Geometry Mathematics 2

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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.

योग

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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Property of an Angle Bisector of a Triangle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Similarity - Practice Set 1.2 [पृष्ठ १५]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
अध्याय 1 Similarity
Practice Set 1.2 | Q 9 | पृष्ठ १५

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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]


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∴ `(square)/("EB") = ("AD")/("DC")`   ...(II) (`square`)

∴ `("AB")/square = square/("EB")`   ...[from (I) and (II)]


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