EuclideanGeometryNotes.pdf

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Chapter 1
Pythagoras Theorem
and Its Applications
1.1 Pythagoras Theorem and its converse
1.1.1 Pythagoras Theorem
The lengths a ≤ b<cof the sides of a right triangle satisfy the relation
a 2 + b 2 = c 2 .
a
b
b
a
c
a
a
c
a
b
c
c
b
b
c
b
a
c
b
a
b
a
1.1.2 Converse Theorem
If the lengths of the sides of a triangles satisfy the relation a 2 +b 2 = c 2 ,then
the triangle contains a right angle.
1
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YIU: Euclidean Geometry
2
A
X
c
b
b
C
a
B
Z
a
Y
Proof. Let ABC be a triangle with BC = a, CA= b,andAB = c satisfy-
ing a 2 + b 2 = c 2 . Consider another triangle XYZ with
YZ= a, XZ = b,
XZY =90
◦
.
By the Pythagorean theorem, XY 2 = a 2 + b 2 = c 2 ,sothatXY = c.
Thus the triangles 4ABC ≡ 4XYZ by the SSS test. This means that
ACB = 6 XZY is a right angle.
Exercise
1. Dissect two given squares into triangles and quadrilaterals and re-
arrange the pieces into a square.
2. BCX and CDY are equilateral triangles inside a rectangle ABCD.
The lines AX and AY are extended to intersect BC and CD respec-
tively at P and Q. Show that
(a) APQ is an equilateral triangle;
(b) 4APB +4ADQ = 4CPQ.
D
Q
C
X
P
Y
A
B
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YIU: Euclidean Geometry
3
3. ABC is a triangle with a right angle at C. If the median on the side
a is the geometric mean of the sides b and c, show that c =3b.
4. (a) Suppose c = a+kb for a right triangle with legs a, b, and hypotenuse
c.Showthat0<k<1, and
a : b : c =1− k 2 :2k :1+k 2 .
(b) Find two right triangles which are not similar, each satisfying c =
3
1
5. ABC is a triangle with a right angle at C.Ifthemedianonthesidec
is the geometric mean of the sides a and b, show that one of the acute
angles is 15
◦
.
6. Let ABC be a right triangle with a right angle at vertex C.Let
CXPY be a square with P on the hypotenuse, and X, Y on the sides.
Show that the length t ofasideofthissquareisgivenby
1
t = 1
b .
a
d
b
b
t
t
a
1/a + 1/b = 1/t.
1/a^2 + 1/b^2 = 1/ d^2.
1 a : b : c =12:35:37or12:5:13. Moregenerally,forh ≤ k,thereis,upto
similarity, a unique right triangle satisfying c = ha + kb provided
(i) h < 1 ≤ k,or
(ii)
4 a + 5 b.
a + 1
√
2
2
≤ h = k<1, or
(iii) h, k > 0, h 2 + k 2 =1.
There are two such right triangles if
0 <h<k<1, 2 + k 2 > 1.
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YIU: Euclidean Geometry
4
7. Let ABC be a right triangle with sides a, b and hypotenuse c.Ifd is
the height of on the hypotenuse, show that
a 2 + 1
b 2 = 1
d 2 .
8. (Construction of integer right triangles) It is known that every right
triangle of integer sides (without common divisor) can be obtained by
choosing two relatively prime positive integers m and n, one odd, one
even, and setting
a = m 2 − n 2 ,
b =2mn,
c = m 2 + n 2 .
(a) Verify that a 2 + b 2 = c 2 .
(b) Complete the following table to fi nd all such right triangles with
sides < 100:
m n a = m 2 − n 2
b =2mn c = m 2 + n 2
(i)
2 1
3
4
5
(ii) 3 2
(iii) 4 1
(iv) 4 3
(v) 5 2
(vi) 5 4
(vii) 6 1
(viii) 6 5
(ix) 7 2
(x) 7 4
(xi) 7 6
(xii) 8 1
(xiii) 8 3
(xiv) 8 5
(xv) 9 2
(xvi) 9 4
65
72
97
1
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YIU: Euclidean Geometry
5
1.2 Euclid’s Proof of Pythagoras Theorem
1.2.1 Euclid’s proof
C
C
C
C
B
A
B
A
B
A
B
A
1.2.2 Application: construction of geometric mean
Construction 1
Given two segments of length a<b,markthreepointsP , A, B on a line
such that PA= a, PB= b,andA, B are on the same side of P . Describe
a semicircle with PB as diameter, and let the perpendicular through A
intersect the semicircle at Q.ThenPQ 2 = PA· PB, so that the length of
PQ is the geometric mean of a and b.
Q
Q
x
P
A
B
P
A
B
PA = a, PB = b; PQ^2 = ab.
a
b
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