MATH 241
Assignment 7
Due Friday March 22 by 12:00 noon
in the assignment box on the 3rd floor of CAB
(1) (6 points) Question 4.7.2: Given 4ABC together with its incircle
meeting the triangle at points D across from A, E across from B
and F across from C, prove that AD BE and CF are concurrent.
This point of concurrency is called the Gergonne point of 4ABC,
in general this point is distinct from the triangle’s incenter.
First, we name the points where AD, BE and CF enter the incircle
X, Y and Z respectively.
A
X
F
E
Y
Z
B
C
D
By Theorem 3.6.2 we have
2
AF = AX·AD = AE 2 ,
BF 2 = BY ·BE = BD2 ,
CD2 = CZ·CF = CE 2
so
AF = AE,
With this we get that
BF = BD,
CD = CE.
AF BD CE
·
·
=1
F B DC EA
and hence the three cevians are concurrent.
1
2
(2) (6 points) Question 4.7.4: Find the length d of segment BD in the
following diagram.
A
2
B
2
C
d
4
E
4
D
F
3
The segment AE is a transversal of 4F BD and so by Menalaus’
Theorem
4 −2 4
F C BA DE
·
·
= ·
· = −1
2 2+d 3
CB AD EF
Thus d = 10/3.
(3) (6 points) Question 4.7.5: Prove that AD is an angle bisector in the
diagram below.
A
4
3
F
E
2
B
1
D
C
Suppose that P is the point on BC so that AP is the angle bisector
of ∠A. Then by the angle bisector theorem BP/P C = AB/AC =
6/4. By Ceva’s Theorem we have
4 BD 1
AF BD CE
·
·
= ·
·
2 DC 3
F B DC EA
So BD/DC = 6/4 also and thus P = D and AD is the angle bisector
of ∠A as required.
1=
3
(4) In the diagram below, find the values of a and b.
3
b
3
a
5
5
4
4
With the vertices labeled as shown,
A
3
F
b
3
a
B
E
G
5
5
D
4
4
C
Menelaus’ Theorem on 4CF B with transversal AD gives
3 −3 5
CG F A BD
·
·
= ·
· .
5 3+a 4
GF AB DC
Solving for a gives a = 13/4. Now by Ceva’s theorem on 4ABC we
have
AF BD CE
3 5 4
1=
·
·
= · · .
a
4 b
F B DC EA
Substituting 13/4 in for a and solving for b gives b = 60/13.
−1 =
4
(5) (6 points) Question 4.7.3: Prove the remaining part of Ceva’s Theorem. That is, prove that if 4ABC has parallel cevians AD BE and
CF then the cevian product equals 1.
For 4ABC to have parallel cevians AD BE and CF exactly one
of the cevians must be internal, without loss of generality, we will
assume that the internal cevian is AD. Let AD, BE and CF be
parallel cevians as shown.
E
F
A
B
D
C
Since E and F are external while D is internal to the triangle sides
we have
AF BD CE
AF BD
CE
AF BD CE
·
·
= (−1)
·
· (−1)
=
·
·
.
F B DC
EA
F B DC EA
F B DC EA
Now, as AD||F C, by Lemma 3.2.1 applied to 4BAD BF/F A =
BC/CD, and hence AF/F B = DC/CB. Similarly, as AD||EB,
CE/EA = CB/BD, so
DC BD CB
AF BD CE
=
·
·
=
·
·
=1
F B DC EA
CB DC BD
Thus, if the cevians AD, BE, CF are parallel, then
AF BD CE
·
·
= 1.
F B DC EA