Question

the proof that ( \triangle qpt cong \triangle qrt ) is shown.
given: ( overline{sp} cong overline{sr} )
line segment ( overline{qt} )
prove: ( \triangle qpt cong \triangle qrt )

Explanation:

Step1: Analyze the statement $\overline{ST}\perp\overline{PR}$

Since $\overline{ST}\perp\overline{PR}$, $\angle STP=\angle STR = 90^{\circ}$. And we know that $\overline{SP}\cong\overline{SR}$ (given) and $\overline{ST}$ is common to both $\triangle SPT$ and $\triangle SRT$.

Step2: Apply the Hypotenuse - Leg (HL) or definition of perpendicular bisector

In right - triangles $\triangle SPT$ and $\triangle SRT$, $\overline{SP}\cong\overline{SR}$ (hypotenuse) and $\overline{ST}\cong\overline{ST}$ (common leg). By the HL congruence theorem for right - triangles (or by the definition of a perpendicular bisector: a point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment), we can conclude that $\overline{PT}\cong\overline{RT}$.

Answer:

definition of perpendicular bisector