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  One of the other perpendicular bisectors must pass through point B

Step-by-step explanation:

Let's look at the choices:

 — incenter equidistant from B and C.

The incenter is on the perpendicular bisector of BC. Every point on that line is equidistant from B and C. This statement is True.

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 — angles B and C are congruent.

As we said above, A (on the perpendicular bisector of BC) is equidistant from B and C. That makes the triangle isosceles. The congruent angles are B and C, opposite congruent sides AC and AB. This statement is True.

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 — the perpendicular bisector of BC passes through the incenter.

The incenter is the point of concurrence of the angle bisectors. Since the perpendicular bisector of BC goes through the a.pex of triangle ABC at A, it is the angle bisector there. Hence the incenter lies on that line. The statement is True.

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 — B lies on one of the other perpendicular bisectors.

This will be true if the triangle is equilateral. Nothing in the problem statement indicates this is the case. The statement is False.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116