We will show that QS is a diameter of the circle L given that QS is a perpendicular bisector of RT.
To do so we will use congruent triangles. Let's consider △ LPT and △ LPR. We are told that L is the center of the circle. Therefore, LT and LR are radii of ⊙ L. Since radii of a circle are congruent, LT and LR are also congruent.
LT≅ LR
Additionally, we can conclude that TP and RP are congruent by the definition of a perpendicular bisector.
TP≅ RP
We can also see that △ LPT and △ LPR share the same leg LP. By the Reflexive Property of Congruence, LP is congruent to itself.
LP≅ LP
Combining all of this information, three sides of △ LPT are congruent to the corresponding three sides of △ LPR.
Therefore, by the Side-Side-Side (SSS) Congruence Theorem △ LPT is congruent to △ LPR.
△ LPT ≅ △ LPR
Since Corresponding Parts of Congruent Triangles are Congruent (CPCTC), we can say that ∠ LPT and ∠ LPR are congruent.
∠ LPT ≅ ∠ LPR
Notice that ∠ TPR is a straight angle. Therefore, we can conclude that ∠ LPT and ∠ LPR are right angles. From here, by the definition of perpendicular bisector LP is the perpendicular bisector of RT.
LP is the ⊥ bisector of RT.
With this information, L lies on QS.
Finally, because QS contains the center of the circle, by the definition of a diameter, QS is a diameter of ⊙ L.
Let's summarize the above process in a flow proof.
Completed Proof
2
&Given:&& QS is a perpendicular
& &&bisector of RT.
&Prove:&& QS is a diameter of the circle L.
Proof:
