Answer :
To prove this theorem, Yori can follow these logical steps:
1. Identify given information:
- Recognize that segment AB is a perpendicular bisector of segment PQ, which means that AB meets PQ at a right angle (90 degrees), and at the point where AB intersects PQ, it divides PQ into two equal parts.
2. Statement of what is to be proven:
- Yori wants to prove that any point X on segment AB is equidistant from points P and Q, meaning that the distance from X to P is the same as the distance from X to Q.
3. Draw a diagram:
- Yori should sketch a picture to visualize the situation. She should have AB intersecting PQ at its midpoint, and she should label the midpoint, say M, where the segments intersect. X is a point on AB selected at random (different from M in general). It's useful to clearly mark the right angles where AB and PQ intersect.
4. Construct triangles and note congruent segments:
- Draw triangles APX and BQX in the diagram by drawing segments from X to P and from X to Q. Because AB bisects PQ, we have PM = MQ. And since AB is a perpendicular bisector, angles APM and BQM are right angles. We know that AX = BX because X lies on AB. Hence, PX and QX are shared by both triangles as common sides.
5. Use congruence postulate:
- Demonstrate that triangle APX is congruent to triangle BQX using the right angle, hypotenuse, side (RHS, sometimes referred to as Hypotenuse-Leg or HL in right triangles) congruence rule.
- We know that triangles are congruent if two pairs of corresponding sides and the angle between them are equal, but in right triangles, if we know the hypotenuse and a leg are equal (which is our case), we can assert that the triangles are congruent.
- In our diagram, angle APM is congruent to angle BQM (both are right angles). PM is congruent to MQ (because AB bisects PQ). And, XM is congruent to itself (reflexive property).
6. Use congruent triangles to show equal lengths:
- By showing that the two triangles APX and BQX are congruent, we know that all corresponding parts of congruent triangles are equal (CPCTC - Corresponding Parts of Congruent Triangles are Congruent). Therefore, AP must be congruent (equal in length) to BQ.
7. Conclude the proof:
- Since Yori has proven that in congruent triangles APX and BQX, sides AP and BQ are equal, it can then be said that any point X on segment AB is equidistant from the endpoints P and Q.
This concludes the proof. Yori has now shown through a series of logical steps that any point on the perpendicular bisector of a segment is equidistant from the endpoints of that segment.
1. Identify given information:
- Recognize that segment AB is a perpendicular bisector of segment PQ, which means that AB meets PQ at a right angle (90 degrees), and at the point where AB intersects PQ, it divides PQ into two equal parts.
2. Statement of what is to be proven:
- Yori wants to prove that any point X on segment AB is equidistant from points P and Q, meaning that the distance from X to P is the same as the distance from X to Q.
3. Draw a diagram:
- Yori should sketch a picture to visualize the situation. She should have AB intersecting PQ at its midpoint, and she should label the midpoint, say M, where the segments intersect. X is a point on AB selected at random (different from M in general). It's useful to clearly mark the right angles where AB and PQ intersect.
4. Construct triangles and note congruent segments:
- Draw triangles APX and BQX in the diagram by drawing segments from X to P and from X to Q. Because AB bisects PQ, we have PM = MQ. And since AB is a perpendicular bisector, angles APM and BQM are right angles. We know that AX = BX because X lies on AB. Hence, PX and QX are shared by both triangles as common sides.
5. Use congruence postulate:
- Demonstrate that triangle APX is congruent to triangle BQX using the right angle, hypotenuse, side (RHS, sometimes referred to as Hypotenuse-Leg or HL in right triangles) congruence rule.
- We know that triangles are congruent if two pairs of corresponding sides and the angle between them are equal, but in right triangles, if we know the hypotenuse and a leg are equal (which is our case), we can assert that the triangles are congruent.
- In our diagram, angle APM is congruent to angle BQM (both are right angles). PM is congruent to MQ (because AB bisects PQ). And, XM is congruent to itself (reflexive property).
6. Use congruent triangles to show equal lengths:
- By showing that the two triangles APX and BQX are congruent, we know that all corresponding parts of congruent triangles are equal (CPCTC - Corresponding Parts of Congruent Triangles are Congruent). Therefore, AP must be congruent (equal in length) to BQ.
7. Conclude the proof:
- Since Yori has proven that in congruent triangles APX and BQX, sides AP and BQ are equal, it can then be said that any point X on segment AB is equidistant from the endpoints P and Q.
This concludes the proof. Yori has now shown through a series of logical steps that any point on the perpendicular bisector of a segment is equidistant from the endpoints of that segment.
