Chapter 10.2, Problem 9PSB

To determine

To prove: that the segment $BC$ is congruence is to the segment $BA$ .

Answer to Problem 9PSB

  $BC\cong BA$ .

Explanation of Solution

Given:

  

Concept used:

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.

If two chords are equidistant from the center, they are congruent.

Calculation:

Perpendicular lines form right angles therefore, the $BEF,BDF$ Right.

Right angles are congruent:

  $BEF\text{ and }BDF$ are congruent.

Bisector bisects an angle equally.

  $FBE\text{ and }FBD$ are congruent.

Reflexive of the line:

  $BF\cong BF$ .

Angle angles side $\left(AAS\right)$ :

Triangle $BFE$ is congruent to triangle $BFD$ .

Corresponding parts of congruent triangles are congruent.

  $DF\cong EF$ .

If two chords are equidistant from the center, they are congruent.

Hence, $BC\cong BA$ .