Question

1. use your knowledge of triangle congruence criteria to write a proof for the following: in the figure, $overline{rx}$ is the perpendicular bisector of $overline{ab}$. prove: $\triangle raxcong\triangle rbx$

Explanation:

Step1: Recall perpendicular - bisector property

Since $\overline{RX}$ is the perpendicular bisector of $\overline{AB}$, we know that $AX = BX$ (by the definition of a perpendicular bisector, it divides a line segment into two equal parts).

Step2: Identify common side

In $\triangle RAX$ and $\triangle RBX$, $\overline{RX}$ is common to both triangles, i.e., $RX=RX$.

Step3: Identify right - angles

$\angle RXA=\angle RXB = 90^{\circ}$ because $\overline{RX}$ is the perpendicular bisector of $\overline{AB}$.

Step4: Apply SAS congruence criterion

In $\triangle RAX$ and $\triangle RBX$, we have $AX = BX$, $\angle RXA=\angle RXB$, and $RX = RX$. By the Side - Angle - Side (SAS) congruence criterion, $\triangle RAX\cong\triangle RBX$.

Answer:

$\triangle RAX\cong\triangle RBX$ (by SAS congruence criterion)