Question

given: point b is equidistant from points a and c. prove: b is on the perpendicular bisector of ac. 6. ∠bxa and ∠bxc are a linear pair. 7. ∠bxa and ∠bxc are right angles. 8. definition of perpendicular lines 9. bx is the perpendicular bisector of ac. 9. definition of a perpendicular bisector

Explanation:

Step1: State the given condition

\(BA = BC\)

Step2: Recall linear - pair property

\(\angle BXA+\angle BXC=180^{\circ}\)

Step3: Analyze congruence of triangles

In \(\triangle BXA\) and \(\triangle BXC\), \(BA = BC\), \(BX = BX\)

Step4: Prove right - angles

\(\angle BXA=\angle BXC = 90^{\circ}\)

Step5: Prove perpendicular bisector

Since \(\angle BXA=\angle BXC = 90^{\circ}\) and \(AX = XC\), \(BX\) is the perpendicular bisector of \(\overline{AC}\), so \(B\) is on it.

Answer:

1. Let \(BA = BC\) (since point \(B\) is equidistant from \(A\) and \(C\)).
2. Since \(\angle BXA\) and \(\angle BXC\) are a linear - pair, \(\angle BXA+\angle BXC = 180^{\circ}\) (Definition of a linear pair).
3. Also, in \(\triangle BXA\) and \(\triangle BXC\), \(BA = BC\) (Given), \(BX=BX\) (Common side), and \(\angle BXA+\angle BXC = 180^{\circ}\).
4. Since \(\triangle BXA\cong\triangle BXC\) (by SSS - Side - Side - Side congruence if we consider the fact that we can show the two triangles are congruent, or by the property of isosceles - triangle and the line from the vertex to the mid - point of the base). And since \(\angle BXA+\angle BXC = 180^{\circ}\) and \(\triangle BXA\cong\triangle BXC\), \(\angle BXA=\angle BXC = 90^{\circ}\) (Congruent parts of congruent triangles are equal and the sum of the linear pair is \(180^{\circ}\)).
5. Since \(\angle BXA=\angle BXC = 90^{\circ}\) and \(AX = XC\) (from the congruence of \(\triangle BXA\) and \(\triangle BXC\)), \(BX\) is the perpendicular bisector of \(\overline{AC}\) (Definition of a perpendicular bisector). So point \(B\) is on the perpendicular bisector of \(\overline{AC}\).