<span>1) We are given that PA = PB, so PA ≅ PB by the definition of the radius.
</span>When you draw a perpendicular to a segment AB, you take the compass, point it at A and draw an arc of size AB, then you do the same pointing the compass on B. Point P will be one of the intersections of those two arcs. Therefore PA and PB correspond to the radii of the arcs, which were taken both equal to AB, therefore they are congruent.
2) We know that angles PCA and PCB are right angles by the definition of perpendicular.
Perpendicularity is the relation between two lines that meet at a right angle. Since we know that PC is perpendicular to AB by construction, ∠PCA and ∠PCB are right angles.
3) PC ≅ PC by the reflexive property congruence.
The reflexive property congruence states that any shape is congruent to itself.
4) So, triangle ACP is congruent to triangle BCP by HL, and AC ≅ BC by CPCTC (corresponding parts of congruent triangles are congruent).
CPCTC states that if two triangles are congruent, then all of the corresponding sides and angles are congruent. Since ΔACP ≡ ΔBCP, then the corresponding sides AC and BC are congruent.
5) Since PC is perpendicular to and bisects AB, P is on the perpendicular bisector of AB by the definition of the perpendicular bisector.
<span>The perpendicular bisector of a segment is a line that cuts the segment into two equal parts (bisector) and that forms with the segment a right angle (perpendicular). Any point on the perpendicular bisector has the same distance from the segment's extremities. PC has exactly the characteristics of a perpendicular bisector of AB. </span>
I hope this helps you
1^4-1^8
1-1
0
Answer: well 2 times 15 = 30
30 times 6 = 180. So the boat is 180 feet long
Step-by-step explanation:
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Which transformations can be used to map a triangle with vertices A(2, 2), B(4, 1), C(4, 5) to A’(–2, –2), B’(–1, –4), C’(–5, –4
Romashka [77]
The triangles ABC and A'B'C' are shown in the diagram below. The transformation is a reflection in the line
. This is proved by the fact that the distance between each corner ABC to the mirror line equals to the distance between the mirror line to A'B'C'.
