Chapter 10.1, Problem 4PSA

To determine

To prove: $\overrightarrow{TQ}\text{ bisects }\angle RTS$ if triangle is inscribed in the circle.

Answer to Problem 4PSA

  $\overrightarrow{TQ}\text{ bisects }\angle RTS$ .

Explanation of Solution

Given:

  

Concept used:

If a radius perpendicular on a chord bisects the chord then triangle form is congruence to each other.

The reflexive property states that for every real number $x,x=x$ .

Corresponding parts of congruent triangles are congruent $\left(CPCTC\right)$

When triangles are congruent, one triangle can be moved to coincide with the other triangle. All corresponding sides and angles will be congruent.

Two triangles are said to be congruent if all the three sides and three angles are same.

Condition of congruence triangle:

Side angle side $\left(SAS\right)$ : two side and one angle of one triangle is equivalent another triangle.

Side-side-side $\left(SSS\right)$ : all three side of the one triangle is equivalent to another triangle.

Angle side angle $\left(ASA\right)$ : two angle and side included between the angles of one triangle is equivalent to another two angle and side included between the angles of triangle.

Angle-angle side $\left(AAS\right)$

two angle and non-included side between the angles of one triangle is equivalent to another two angle and non-included side between the angles of triangle.

Right angle hypotenuse side $\left(RHS\right)$: Hypotenuse and side of a right-angled triangle is equivalent to the hypotenuse and side of the second right angled triangle.

Calculation:

According to the given:

  $\overline{QT}\perp \overline{RS}$ .

Since, a radius perpendicular on a chord bisects the chord.

  $\overline{PS}\cong \overline{PR}$ .

According to the reflexive property:

  $\overline{TP}\cong \overline{TP}$ .

Therefore,

  $\Delta TPS\cong TPR$ .

Since, congruent parts of congruent triangles are congruent.

  $\angle STP\cong \angle RTP$ .

By definition of angle bisector:

  $\overrightarrow{TQ}\text{ bisects }\angle RTS$ .

Hence, $\overrightarrow{TQ}\text{ bisects }\angle RTS$ .