3.02 Quiz: Constructions of Segments Angles and Bisectors What are the steps for using a compass and straightedge to construct the bisector of ZR?
R
Drag the steps and drop them in order from start to finish.
Use the straightedge to draw RU
Without changing the opening of the compass, place the point of the compass on point T and draw another are in the interior of
the angle.
Place the point of the compass on point S and draw an arc in the interior of the angle.
Place the point of the compass on point R and draw an arc that intersects the sides of ZR. Label the points of intersection as

The steps to construct the bisector of ∠R using a compass and a straightedge in order from start to finish are;

1) Place the point of the compass on point R and draw an arc that intersects the sides of ∠R. Label the points of intersection as T and S

2) Place the point of the compass on point S and draw an arc in the interior of the angle

3) Without changing the opening of the compass, place the point of the compass on the point T and draw another arc in the interior of angle ∠R. Label the point of intersection of the two arcs as U

4) Use the straightedge to draw RU

Step-by-step explanation:

In order to construct the bisector to an angle

The compass is used to draw an arc that intersects the two rays forming the angle.

From the point of intersection of the arc constructed above and the two rays, two arcs are drawn with the same radius, on the inside of the angle to be bisected, intersecting each other

From the point of intersection of the two arcs drawn wit the same radius, drawn from the intersection of the first arc and the two rays forming the angle to be bisected, a line is drawn with a rule or straightedge to the vertex of the angle to be bisected.

4

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)