Chapter 10.2, Problem 8PSB

To determine

To prove: that the segment $\text{MQ}$ is congruence to the segment $QN$ .

Answer to Problem 8PSB

  $\text{MQ }\cong QN.$

Explanation of Solution

Given:

  

Concept used:

Perpendicular from center to chord bisects the chord.

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:

  $PS\perp RT$ .

Which implies $RS=ST$ are perpendicular from center to chord bisects the chord.

Consider $\Delta RSP\text{ and }\Delta TSP$ :

  $\begin{array}{l}RS=ST.\\ \angle RSP=\angle TSP.\end{array}$

  $SP$ is a common side.

  $\Delta RSP$ is congruent to the triangle $\Delta TSP$ by $SAS$ congruence criterion.

Since, congruent parts of congruent triangles are congruent.

  $RP=TP.$

According to the given:

  $\begin{array}{l}MQ\perp RP\text{ and }NQ\perp TP.\\ \Rightarrow MP=NP.\end{array}$

As it is perpendicular from center to chord bisects the chord.

Consider right triangles $MQP\text{ and }NQP$ :

  $QP$ common hypotenuse.

  $\text{Leg }MP=NP$ .

By $HL$ postulate $MQP$ is congruent to $NQP$ .

Since, congruent parts of congruent triangles are congruent.

Hence, $\text{MQ }\cong QN.$