Solutions Manual
Classical Geometry
Euclidean, Transformational, Inversive,
and Projective
By
I. E. Leonard,
G. W. Tokarsky,
A. C. F. Liu,
J. E. Lewis
( All Chapters Included - 100% Verified Solutions )
1
,CHAPTER 1
CONGRUENCY
1. Prove that the internal and external bisectors of the angles of a triangle are
perpendicular.
Solution. Let BD and BE be the angle bisectors, as shown in the diagram
below.
A
Then
Z.EBD = ZEBA+ZDBA = = ^ A + CBA = ^
Solutions Manual to Accompany Classical Geometry: Euclidean, Transformational, Inversive, and 3
Projective, First Edition. By I. E. Leonard, J. E. Lewis, A. C. F. Liu, G. W. Tokarsky.
Copyright © 2014 John Wiley & Sons, Inc. Published 2014 by John Wiley & Sons, Inc.
2
,4 CONCURRENCY
3. Let P be a point inside A ABC. Use the Triangle Inequality to prove that
AB + BC > AP + PC.
Solution. Extend AP to meet BC at D. Using
the Triangle Inequality,
AB + BD > AD = AP 4- PD
so that
AB + BD + DC > AP + PD + DC.
Since
BD + DC = BC
and
PD -f DC > PC,
we have
AB + BC > AP + PC.
5. Given the isosceles triangle ABC with AB = AC, let D be the foot of the
perpendicular from A to BC. Prove that AD bisects ABAC.
Solution. Referring to the diagram, the two
right triangles ADB and ADC have a com-
mon side and equal hypotenuses, so they are
congruent by HSR. Consequently, ZBAD =
ACAD.
1. D is a point on BC such that AD is the bisector of Z.A. Show that
/.ADC = 90 +
3
, 5
Solution. Referring to the diagram, 29 + (3 +
7 = 180, which implies that
A
From the Exterior Angle Theorem, we have
so that B C
D
9. Construct a right triangle given the hypotenuse and one side.
Solution. We construct a right triangle ABC given the hypotenuse BC and
the length c of side AC.
Construction.
(1) Construct the right bisector of BC, yielding M, the midpoint of BC.
(2) With center M, draw a semicircle with diameter BC.
(3) With center C and radius equal to c, draw an arc cutting the semicircle at
A.
Then ABC is the desired triangle.
C M B
A
Justification. LB AC is a right angle by Thales' Theorem.
11. Let Q be the foot of the perpendicular from a point P to a line I. Show that Q
is the point on I that is closest to P.
Solution. Let X be any point on I with X ^ Q, as in the figure below.
P
4
Classical Geometry
Euclidean, Transformational, Inversive,
and Projective
By
I. E. Leonard,
G. W. Tokarsky,
A. C. F. Liu,
J. E. Lewis
( All Chapters Included - 100% Verified Solutions )
1
,CHAPTER 1
CONGRUENCY
1. Prove that the internal and external bisectors of the angles of a triangle are
perpendicular.
Solution. Let BD and BE be the angle bisectors, as shown in the diagram
below.
A
Then
Z.EBD = ZEBA+ZDBA = = ^ A + CBA = ^
Solutions Manual to Accompany Classical Geometry: Euclidean, Transformational, Inversive, and 3
Projective, First Edition. By I. E. Leonard, J. E. Lewis, A. C. F. Liu, G. W. Tokarsky.
Copyright © 2014 John Wiley & Sons, Inc. Published 2014 by John Wiley & Sons, Inc.
2
,4 CONCURRENCY
3. Let P be a point inside A ABC. Use the Triangle Inequality to prove that
AB + BC > AP + PC.
Solution. Extend AP to meet BC at D. Using
the Triangle Inequality,
AB + BD > AD = AP 4- PD
so that
AB + BD + DC > AP + PD + DC.
Since
BD + DC = BC
and
PD -f DC > PC,
we have
AB + BC > AP + PC.
5. Given the isosceles triangle ABC with AB = AC, let D be the foot of the
perpendicular from A to BC. Prove that AD bisects ABAC.
Solution. Referring to the diagram, the two
right triangles ADB and ADC have a com-
mon side and equal hypotenuses, so they are
congruent by HSR. Consequently, ZBAD =
ACAD.
1. D is a point on BC such that AD is the bisector of Z.A. Show that
/.ADC = 90 +
3
, 5
Solution. Referring to the diagram, 29 + (3 +
7 = 180, which implies that
A
From the Exterior Angle Theorem, we have
so that B C
D
9. Construct a right triangle given the hypotenuse and one side.
Solution. We construct a right triangle ABC given the hypotenuse BC and
the length c of side AC.
Construction.
(1) Construct the right bisector of BC, yielding M, the midpoint of BC.
(2) With center M, draw a semicircle with diameter BC.
(3) With center C and radius equal to c, draw an arc cutting the semicircle at
A.
Then ABC is the desired triangle.
C M B
A
Justification. LB AC is a right angle by Thales' Theorem.
11. Let Q be the foot of the perpendicular from a point P to a line I. Show that Q
is the point on I that is closest to P.
Solution. Let X be any point on I with X ^ Q, as in the figure below.
P
4
