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Prove by Vector Method that the Internal Bisectors of the Angles of a Triangle Are Concurrent. - Mathematics

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प्रश्न

Prove by vector method that the internal bisectors of the angles of a triangle are concurrent.

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उत्तर


Let ABC be a triangle and \[\vec{\alpha} , \vec{\beta} ,\vec{\gamma}\] be the position vectors of the vertices  A, B and C respectively.
Let AD, BEand CF be the internal bisectors of \[\angle A, \angle B\] and \[\angle C\] respectively.
We know that D divides BC in the ratio of AB : AC that is c : b.
Then,
P.V. of D is \[\frac{c \vec{\gamma} + b \vec{\beta}}{c + b}\].
P.V. of E is\[\frac{c \vec{\gamma} + a \vec{\alpha}}{c + a}\] and 
P.V. of F is \[\frac{a \vec{\alpha} + b \vec{\beta}}{a + b}\]
The point dividing AD in the ratio \[b + c : a\] is \[\frac{a \vec{\alpha} + b \vec{\beta} + c \vec{\gamma}}{a + b + c}\]
The point dividing BE in the ratio of \[a + c : b\] is\[\frac{a \vec{\alpha} + b \vec{\beta} + c \vec{\gamma}}{a + b + c}\]
The point dividing CF in the ratio of a + b : c is \[\frac{a \vec{\alpha} + b \vec{\beta} + c \vec{\gamma}}{a + b + c}\]
Since the point \[\frac{a \vec{\alpha} + b \vec{\beta} + c \vec{\gamma}}{a + b + c}\]  lies on all the three internal bisectors AD, BE and CF.
Hence the internal bisectors are concurrent .

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Position Vector of a Point Dividing a Line Segment in a Given Ratio
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पाठ 23: Algebra of Vectors - Exercise 23.4 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 23 Algebra of Vectors
Exercise 23.4 | Q 6 | पृष्ठ ३७

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