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(A) Let the vectors $a$ and $b$ be represented by sides $\vec{OA}$ and $\vec{OD}$ of a parallelogram $OADC$. The diagonal $\vec{OC} = a+b$ bisects the

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$|a|=|b|$

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(A) Let the vectors $a$ and $b$ be represented by sides $\vec{OA}$ and $\vec{OD}$ of a parallelogram $OADC$. The diagonal $\vec{OC} = a+b$ bisects the angle between $a$ and $b$. Let the angle between $a$ and $b$ be $2	heta$. Then the angle between $a$ and $a+b$ is $	heta$. In the parallelogram $OADC$,we have $OA = |a|$ and $OD = AC = |b|$. Consider the triangle $	riangle ABC$ formed by dropping a perpendicular from $C$ to the line $OA$ extended to $B$. In $	riangle ABC$,$AB = |b| \cos 2	heta$ and $BC = |b| \sin 2	heta$. In the right-angled triangle $	riangle OBC$,$	an 	heta = \frac{BC}{OB} = \frac{|b| \sin 2	heta}{|a| + |b| \cos 2	heta}$. Using the double angle formulas $\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$ and $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$,we get: $	an 	heta = \frac{|b| \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight)}{|a| + |b| \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight)}$ $	an 	heta = \frac{2|b| 	an 	heta}{|a|(1 + 	an^2 	heta) + |b|(1 - 	an^2 	heta)}$ Since $a$ and $b$ are non-collinear,$	heta 
eq 0$,so $	an 	heta 
eq 0$. Dividing by $	an 	heta$: $1 = \frac{2|b|}{|a| + |a| 	an^2 	heta + |b| - |b| 	an^2 	heta}$ $|a| + |a| 	an^2 	heta + |b| - |b| 	an^2 	heta = 2|b|$ $|a|(1 + 	an^2 	heta) - |b|(1 + 	an^2 	heta) = 0$ $(|a| - |b|)(1 + 	an^2 	heta) = 0$ Since $1 + 	an^2 	heta = \sec^2 	heta 
eq 0$,we must have $|a| - |b| = 0$,which implies $|a| = |b|$.

If $a$ and $b$ are two non-collinear vectors and the vector $a+b$ bisects the angle between $a$ and $b$,then

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