Chapter 10.1, Problem 15PSB

To determine

To prove: that segment $\overline{QR}$ is congruence to the $\overline{TS}$ if segment $\overleftrightarrow{QT}$ is parallel to the $\overleftrightarrow{RS}$ .

Answer to Problem 15PSB

  $\overline{QR}\cong \overline{TS}$ .

Explanation of Solution

Given:

  

Concept used:

The side splitter theorem states that if a line is parallel to aside of a triangle and the line intersects the other two sides, then this line divides those two sides proportionally.

The radii of circle equidistant from the circumference therefore, the radii of the circle are congruent.

Substitution method: by substituting the equal quantity in the equation or function.

Calculation:

According to the given:

  $\overleftrightarrow{QT}\parallel \overleftrightarrow{RS}$ .

By using side splitter theorem:

  $\frac{PQ}{QR}=\frac{PT}{TS}$ .

The radii of a circle are congruent.

  $PQ=PT$ .

By substitution method:

  $\begin{array}{l}\frac{PQ}{QR}=\frac{PT}{TS}.\\ QR=TS.\end{array}$ .

Hence, $\overline{QR}\cong \overline{TS}$ .