If a, b, c are lengths of the sides BC, CA, A | vedclass

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(B) Let the position vector of $H$ be the origin $\vec{0}$.Then the position vectors of the vertices $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ respect

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Incentre of $	riangle ABC$

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(B) Let the position vector of $H$ be the origin $\vec{0}$.Then the position vectors of the vertices $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ respectively.The given equation is $a \vec{AH} + b \vec{BH} + c \vec{CH} = \vec{0}$.Since $H$ is the origin,$\vec{AH} = \vec{0} - \vec{a} = -\vec{a}$,$\vec{BH} = -\vec{b}$,and $\vec{CH} = -\vec{c}$.Substituting these into the equation,we get $a(-\vec{a}) + b(-\vec{b}) + c(-\vec{c}) = \vec{0}$,which implies $a\vec{a} + b\vec{b} + c\vec{c} = \vec{0}$.However,the standard definition of the incentre $I$ with position vector $\vec{i}$ is $\vec{i} = \frac{a\vec{A} + b\vec{B} + c\vec{C}}{a+b+c}$.If $H$ is the incentre,then $a(\vec{A}-\vec{H}) + b(\vec{B}-\vec{H}) + c(\vec{C}-\vec{H}) = \vec{0}$.This simplifies to $(a+b+c)\vec{H} = a\vec{A} + b\vec{B} + c\vec{C}$,which is the definition of the incentre.Therefore,$H$ is the incentre of $	riangle ABC$.

If $a, b, c$ are lengths of the sides $BC, CA, AB$ respectively of $\triangle ABC$ and $H$ is any point in the plane of $\triangle ABC$ such that $a \vec{AH} + b \vec{BH} + c \vec{CH} = \vec{0}$,then $H$ is the

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