Question

given: point b is equidistant from points a and c. prove: b is on the perpendicular bisector of ac. 1. ab ≅ bc 1. given 2. draw median bx, ax ≅ xc 2. definition of a median 3. bx ≅ bx 3. 4. △abx ≅ △cbx 4. 5. 5. corresponding parts of congruent triangles are congruent

Explanation:

Step1: Justify $\overline{BX}\cong\overline{BX}$

By the reflexive property of congruence, any segment is congruent to itself.

Step2: Justify $\triangle ABX\cong\triangle CBX$

We know $\overline{AB}\cong\overline{BC}$ (given), $\overline{AX}\cong\overline{XC}$ (by definition of median) and $\overline{BX}\cong\overline{BX}$ (reflexive property). So by SSS congruence postulate, $\triangle ABX\cong\triangle CBX$.

Step3: Prove $\overline{BX}$ is the perpendicular bisector of $\overline{AC}$

Since $\triangle ABX\cong\triangle CBX$, by corresponding parts of congruent triangles are congruent, $\angle BXA\cong\angle BXC$. Also, $\angle BXA+\angle BXC = 180^{\circ}$ (linear - pair of angles). So $\angle BXA=\angle BXC = 90^{\circ}$, which means $\overline{BX}\perp\overline{AC}$. And we already know $\overline{AX}\cong\overline{XC}$ from the definition of the median. So $\overline{BX}$ is the perpendicular bisector of $\overline{AC}$, and point $B$ lies on it.

Answer:

1. Reflexive Property of Congruence; 4. SSS (Side - Side - Side) Congruence Postulate; 5. $\angle BXA\cong\angle BXC$, $\overline{BX}\perp\overline{AC}$ and $\overline{AX}\cong\overline{XC}$ so $\overline{BX}$ is the perpendicular bisector of $\overline{AC}$