Section
8.1
217
3. (a) [BB]
Let
A
be any point on
f
and let
r
=
IAPI.
Draw the circle with center
P
and radius
r.
If
this meets
f
in only the single point
A,
then
PA
is the desired perpendicular to
f.
Otherwise, the
circle meets
f
in two points,
A
and
B.
Draw arcs with centers
A
and
B
and radius
r
meeting in
P
4.
and
Q.
Then
PQ
is the desired perpendicular to
f.
'\Pi
To see this, consider the diagram at the right,
with
M
the point
of
intersection
of
PQ
and
f.
Since triangles
P
AQ
and
P
BQ
are
.........
4
congruent,
LAPM
=
LBPM.
Then it
follows that
.6.APM
==
.6.BP M
and thus,
LAMP
=
LBMP.
Since these angles have
sum 180
0
,
each is a right angle.
I.J
.........
f
(b) Draw an arc with center
P
and any radius
r
and let
A
and
B
be
the points
of
intersection
of
this
arc with
f.
Then draw arcs with centers
A
and
B
and any radius
s
>
r.
Letting
C
be one
of
the points
of
intersection
of
these
\r:l
arcs,
CP
is the required perpendicular. To see this,
note that triangles
CPA
and
C P B
have three pairs
of
equal sides and so are congruent. Thus,
y1
.fJ'
f
LC
P A
=
LC
P B
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 Summer '10
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 Graph Theory, 2k, russian peasant method

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