Converse Perpendicular Chord Bisector Theorem

It is given that $\overline{Q S}$ is a perpendicular bisector of $\overline{T R} .$ Let $L$ be the center of the circle. Consider $△ L P T$ and $△ L P R .$

Segments

\overline{L T}

and

\overline{L R}

are radii of

⊙ L .

Since all radii of a circle are congruent,

\overline{L T}

and

\overline{L R}

are congruent. Additionally, by the definition of a perpendicular bisector, it can be stated

\overline{T P}

and

\overline{R P}

are congruent segments. Furthermore, by the Reflexive Property of Congruence, it can be said that

\overline{L P}

is congruent to itself.

\overline{L T} ≅ \overline{L R} \overline{T P} ≅ \overline{R P} \overline{L P} ≅ \overline{L P}

This information can be seen in the diagram.

Here, three sides of

△ L P T

are congruent to three sides of

△ L P R .

Therefore, by the Side-Side-Side Congruence Theorem,

△ L P T

and

△ L P R

are congruent triangles. Since corresponding parts of congruent triangles are congruent,

∠ L P T

and

∠ L P R

are congruent angles.

△ L P T ≅ △ L P R ⇓ ∠ L P T ≅ ∠ L P R

By the definition of a chord,

∠ T P R

is a straight angle. By dividing

18 0^{\circ}

2,

it can be calculated that

∠ L P T

and

∠ L P R

measure

9 0^{\circ}

each.

m ∠ L P T = 9 0^{\circ} and m ∠ L P R = 9 0^{\circ}

This means that they are right angles and that

\overline{L P}

is perpendicular to

\overline{T R} .

Therefore,

\overline{L P}

is the perpendicular bisector of

\overline{R T} .

Since there is only one perpendicular bisector that can be drawn to a segment,

L

must lie on

\overline{Q S} .

Finally, the fact that $\overline{Q S}$ contains the center of the circle leads to the conclusion that $\overline{Q S}$ is a diameter of $⊙ L .$

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Converse Perpendicular Chord Bisector Theorem

Proof

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