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A triangle \(ABC\) has a fixed base \(BC\). If \(AB:AC= 1:2\), then prove that the locus of the vertex \(A\) is a circle whose centre is on the side \(BC\) but not the midpoint of \(BC\).

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See http://www.cut-the-knot.org/Curriculum/Geometry/LocusCircle.shtml

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Thanks!

Assume the side BC lies on X axis with the mid point as Origin (0,0) and B (-a,0) C (a,0). Let the locus point A be (x,y)

Now since AB: AC = 1:2

\frac { \sqrt { { (x+a) }^{ 2 }+{ y }^{ 2 } } }{ \sqrt { { (x-a) }^{ 2 }+{ y }^{ 2 } } } =\frac { 1 }{ 2 }

Solving this you'll get a circle equation proving that the locus of A is a circle with centre on BC i.e on X axis

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TopNewestSee http://www.cut-the-knot.org/Curriculum/Geometry/LocusCircle.shtml

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Thanks!

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Assume the side

BClies on X axis with the mid point as Origin (0,0) and B (-a,0) C (a,0). Let the locus point A be (x,y)Now since AB: AC = 1:2

\frac { \sqrt { { (x+a) }^{ 2 }+{ y }^{ 2 } } }{ \sqrt { { (x-a) }^{ 2 }+{ y }^{ 2 } } } =\frac { 1 }{ 2 }

Solving this you'll get a circle equation proving that the locus of A is a circle with centre on BC i.e on X axis

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Nice...

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