what is the first step when constructing an angle bisector using only a compass and a straightedge?

a. place the compass needle on one of the legs of the angle, and draw an arc intersecting the other leg.
b. mark a point outside the angle, and draw an arc centered at the point, intersecting the vertex of the angle.
c. mark a point outside the angle, and draw a ray in any direction from the point.
d. place the compass needle on the vertex of the angle, and draw an arc across both legs of the angle.
e. mark a point in the angle's interior, and draw an arc centered at the point, intersecting both legs of the angle.

D

Step-by-step explanation:

Then you find the intersection and get the bisector without a protractor :)

D

Step-by-step explanation:

To answer this question, we have to define to bisect an angle:

To bisect an angle is to divide it into two equal parts, we can call those as congruent parts because these have the same area.

So it is crucial: we need to keep it proportionally divided when we start to construct an angle bisector.

Examining the alternatives, there is only one correct:

--

D. So now let's place the compass needle on the vertex of the angle, and draw an arc across both legs of the angle. True

The reason to proceed this way as our first step is the proportional division made by the inner line that bisects the angle. Thus in the first step, we pave the way to trace another arc in the middle of the inside area. Done with the same width and compass needle on the marks. Finally, the line to vertex crossing the mark.

(Check the 3rd picture to Final Construction)

after simplifying it i got: -36.25 as my answer. hope that !

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.