Draw any obtuse angle.Bisect it using compass
Answers
Given angle
∠
A
B
C
with vertex
B
and two sides
B
A
and
B
C
. It can be acute or obtuse, or right - makes no difference.
Choose any segment of some length
d
and mark point
M
on side
B
A
on a distance
d
from vertex
B
.
Using the same segment of length
d
, mark point
N
on side
B
C
on distance
d
from vertex
B
.
Red arc on a picture represents this process, its ends are
M
and
N
.
We can say now that
B
M
≅
B
N
.
Choose a radius sufficiently large (greater than half the distance between points
M
and
N
) and draw two circles with centers at points
M
and
N
of this radius. These two circles intersect in two points,
P
and
Q
. See two small arcs intersecting on a picture, their intersection is point
P
.
Chose any of these intersection points, say
P
, and connect it with vertex
B
. This is a bisector of an angle
∠
A
B
C
.
Proof
Compare triangles
Δ
B
M
P
and
Δ
B
N
P
.
1. They share side
B
P
2.
B
M
≅
B
N
by construction, since we used the same length
d
to mark both points
M
and
N
3.
M
P
≅
N
P
by construction, since we used the same radius of two intersecting circles with centers at points
M
and
N
.
Therefore, triangles
Δ
B
M
P
and
Δ
B
N
P
are congruent by three sides:
Δ
B
M
P
≅
Δ
B
N
P
As a consequence of congruence of these triangles, corresponding angles have the same measure.
Angles
∠
M
B
P
and
∠
N
B
P
lie across congruent sides
M
P
and
N
P
.
Therefore, these angles are congruent:
∠
M
B
P
≅
∠
N
B
P
,
that is
B
P
is a bisector of angle
∠
M
B
P
(which is the same as angle
∠
A
B
C
).
Steps of construction:
Step 1: Draw an obtuse angle. We choose ∠ABC=120
0
.
Step 2: Draw an arc which intersects AB at P and BC at Q, from center B and choose any radius.
Step 3: Draw an arc from point P by setting radius more than half of PQ.
Step 4: Repeat step 3 using Q as center and cut the previous arc at R.
Step 5: Join BR.
The diagram is shown in the above image.
Therefore, ∠ABR=∠RBC=60
0