Download Formulae Connecting Segments of the Same Line Pure Geometry

Pure Geometry
CHAPTER
1
Formulae Connecting
Segments of the Same Line
P
1. One of the differences between Modern Geometry and the Geometry of Euclid
is that a length in Modern Geometry has a sign as well as a magnitude. Lengths
measured on a line in one direction are considered positive and those measured in
the opposite direction are considered negative. Thus if AB, i.e., the segment extending
from A to B, be considered positive, then BA, i.e., the segment extending from B to
A, must be considered negative. Also AB and BA differ only in sign. Hence we obtain
the first formula, viz. AB = – BA.
Notice that by allowing lengths to have a sign as well as a magnitude, we are
enabled to utilize the formulae of Algebra in geometrical investigations. In making
use of Algebra it is generally best to reduce all the segments we employ to the same
origin. This is done in the following way.
O
A
B
Take any segment AB on a line and also any origin O. Then AB = OB – OA. This
is obviously true in the above figure, and it is true for any figure. For
OB – OA = OB + AO = AO + OB = AB;
for AO + OB means that the point travels from A to O and then from O to B, and
thus the point has gone from A to B.
The fundamental formulae then are
(1) AB = – BA;
(2) AB = OB – OA.
In the above discussion the lengths have been taken on a line. But this is not
necessary; the lengths might have been taken on any curve.
It is generally convenient to use an abridged form of the formula AB = OB – OA,
viz. AB = b – a, where a = OA and b = OB.
P 2. A, B, C, D are any four collinear points; show that
AB. CD + AC . DB + AD . BC = 0.
Take A as origin, then CD = AD – AC = d – c, and so on.
Hence
AB . CD + AC . DB + AD . BC = b(d – c) + c(b – d) + d(c – b)
= bd – bc + cb – cd + dc – db = 0.
1
2
AN ELEMENTARY TREATISE ON PURE GEOMETRY WITH NUMEROUS EXAMPLES
EXAMPLE 1. A, B, C, D, O are any five points in a plane; show that
Δ AOB . Δ COD + Δ AOC . Δ DOB + Δ AOD . Δ BOC = 0,
where Δ AOB denotes the area of the triangle AOB.
Let a line meet OA, OB, OC, OD in A′, B′, C′, D′. Then
1
Δ AOB = OA . OB sin AOB.
2
Hence the given relation is true if
Σ{sin AOB . sin COD} = 0,
i.e. if Σ{sin A′OB′ . sin C′OD′} = 0.
But p. A′B′ = OA′ . OB′ sin A′OB′, where p is the perpendicular from O on A′B′C′D′.
Hence the given relation is true if A′B′. C′D′ + A′C′. D′B′ + A′D′. B′C′ = 0.
EXAMPLE 2. If OA, OB, OC, OD be any four lines meeting in a point, show that
sin AOB. sin COD + sin AOC. sin DOB + sin AOD. sin BOC = 0.
EXAMPLE 3. From Ex. 2 deduce Ptolemy’s Theorem connecting four points on a
circle.
Take O also on the circle. Then AB = 2. R. sin AOB.
EXAMPLE 4. If O, A, B are any collinear points, then
OA2 + OB2 = AB2 + 2OA. OB.
EXAMPLE 5. If A, B, C, ... X, Y are any number of collinear points, show that
AB + BC+ ... + XY + YA = 0.
EXAMPLE 6. If λ denotes the ratio OA : OB and λ′ the ratio OA′: OB′ (O, A, B, A′,
B′ being collinear points), show that
BB′. λ . λ′ + A′ B. λ + B′A. λ′ + AA′ = 0.
EXAMPLE 7. If D, E, F are any three points on the sides BC, CA, AB of a triangle,
show that
DB . EC . FA
sin DAB . sin EBC . sin FCA
=
DC . EA . FB
sin DAC . sin EBA . sin FCB.
STEWART’S THEOREM
P
3. A, B, C are any three collinear points, and P is any other point; show that
PA2 . BC + PB2 . CA + PC2 . AB + BC . CA . AB = 0.
Drop the perpendicular PO from P on ABC.
Take A as origin (so that the factors CA and AB may be obvious);
and let
AO = x and PO = p.
Then
PA2 . BC + PB2 . CA + PC2 . AB
= (OA2 + OP2) BC + (OB2 + OP2) CA + (OC2 + OP2) AB
= (x2 + p2) (c – b) + [(b – x)2 + p2] (– c) + [(c – x)2 + p2]b
= x2c – x2b + (b2 – 2bx + x2) (– c) + (c2 – 2cx + x2)b
+ p2 (c – b – c + b)
PURE GEOMETRY
3
= – b2c + c2b = bc (c – b) = AB . AC . BC
= – BC . CA . AB.
EXAMPLE 1. If A, B, C be three collinear points and a, b, c be the tangents from
A, B, C to a given circle, then
a2 . BC + b2 . CA + c2 . AB + BC . CA . AB = 0.
E XAMPLE 2. If P be any point on the base AB of the triangle ABC, then
AP . CB2 – BP . CA2 = AB . (CP2 – AP . BP).
EXAMPLE 3. If A, B, C, D be four points on a circle and P any point, show that
Δ BCD . AP2 – ΔCDA . BP2 + Δ DAB . CP2 – Δ ABC . DP2 = 0,
disregarding signs.
Let AC, BD meet in O inside the circle.
Then
Δ BCD ∝ BD . CO and BO . OD = CO . OA.
EXAMPLE 4. If VA, VB, VC, VD be any four lines through V, then
sin AVD . sin BVD
sin BVD . sin CVD
sin CVD . sin AVD
+
+
= 1.
sin AVC . sin BVC
sin BVA . sin CVA
sin CVB . sin AVB
Draw a parallel to VD.
EXAMPLE 5. If from any point P there be drawn the perpendicular PQ on the line
AB, then
PA2 – PB2 = AB2 + 2 . AB . BQ.
P 4. If C be the middle point of AB, then whatever origin O be chosen on the line
1
AB, we have OC = (OA + OB).
2
1
1
For
OC = OA + AC = OA + AB = OA + (OB – OA)
2
2
1
= (OA + OB).
2
As we have used general formulae throughout this proof, the formula holds for
every relative position of the points O, A, and B.
EXAMPLE 1. If A A′, BB′, CC′ be collinear segments whose middle points are α, β,
γ, and if P be a variable point on the line, show that
PA . PA′ . βγ + PB . PB′ . γ α + PC . PC′. αβ is constant.
Take any origin O. Then
2 . βγ = 2 . Oγ – 2 . Oβ = c + c′ – b – b′.
Twice the given expression is
(d – p) (a′ – p) (c + c′ – b – b′) + ... + ...,
which is equal to aa′ (c + c′ – b – b′) + ... + ...
EXAMPLE 2. If C be the middle point of AB, and O be any point on the line ACB,
show that
OA2 + OB2 = CA2 + CB2 + 2 . OC2.
EXAMPLE 3. If P be the middle point of the segment A A′ and Q be the middle point
of the segment BB′ (on the same line as AA′), show that
2 . PQ . AA′ = AB . AB′ – A′B . A′B′.
4
AN ELEMENTARY TREATISE ON PURE GEOMETRY WITH NUMEROUS EXAMPLES
EXAMPLE 4. If on the line AB the point G be taken such that a . GA + b . GB = 0,
a and b being any numbers, positive or negative, then, O being also on AB,
a . OA + b . OB = (a + b) . OG.
EXAMPLE 5. If on the line ABCD ... a point G be taken such that
GA + GB + GC + ... = 0,
and O be any other point on the line, then
OA2 + OB2 + OC2 + ... = GA2 + GB2 + GC2 + ... + n . GO2, n being the number of
the points ABCD ....
P 5. The following is an interesting application of Algebra to Geometry:
If A, B, C, D, P, Q be any six collinear points, then
DP . DQ
CP . CQ
BP . BQ
AP . AQ
+
+
+
= 0.
DA . DB . DC
CD . CA . CB
BC . BD . BA
AB . AC . AD
Put X for A, and reduce the resulting equation to any origin, after getting rid of
the denominators. We shall have an equation of the second order in x to determine X.
Put
x = b, i.e. X = B, and we get an identity.
Hence
x = b is one solution of this equation.
Similarly
x = c, and x = d are solutions.
Hence the equation of the second order has three solutions; and is therefore an
identity.
If A, B, C, P, Q be any five collinear points, then
CP . CQ
AP . AQ
BP . BQ
+
+
= 1.
CA . CB
AB . AC
BC . BA
Multiply the identity just proved by AD throughout and let D be at infinity.
Then
AD = AB + BD, ∴ AD/BD = AB/BD + 1.
But when D is at infinity
AB/BD = 0.
Hence
AD/BD = 1.
Similarly AD/CD = 1.
so
DP/DB = 1
and DQ/DC = 1.
Hence we obtain the result enunciated.
If A, B, C, D, P be any five collinear points, then
AP
BP
CP
DP
+
+
+
= 0.
AB . AC . AD BC . BD . BA CD . CA . CB DA . DB . DC
In the first identity take Q at infinity, then since
BQ/AQ = 1, CQ/AQ = 1, DQ/AQ = 1,
the required result follows.
EXAMPLE 1. Show that the first result is true for n points A, B, ... and (n – 2) points
P, Q, ....
EXAMPLE 2. Show that the second result is true for n points A, B, ... and (n – 1)
points P, Q, ....
EXAMPLE 3. Show that the third result is true for n points A, B, ... and (n – 2 – m)
points P, Q, ..., where m may be 0, 1, 2, 3, ... (n – 2).
PURE GEOMETRY
5
EXAMPLE 4. Enunciate the theorems obtained from Ex. 2 and Ex. 3 by taking the
points P, Q, ... all coincident; and show that the theorems still hold when P is outside
the line, provided the index of AP is even.
Use
AP2 = Ap2 + pP2, and the Binomial Theorem.
MENELAUS’S THEOREM
P
6. If any transversal meet the sides BC, CA, AB of a triangle in D, E, F, then
AF . BD . CE = – FB . DC . EA.
The transversal must cut all the sides externally, or two sides internally and one
externally; for as a point proceeds along the transversal from infinity, at a point where
the transversal cuts a side internally, the point enters the triangle and at the point
where the point leaves the triangle, the transversal must cut another side internally.
Hence of the ratios AF : FB, BD : DC, CE : EA, one is negative and the other two
are either both positive or both negative. Hence the sign of the formula is correct.
E
A
F
F
B
D
C
D
E
To prove that the formula is numerically correct, drop the perpendiculars p, q, r
from A, B, C on the transversal. Then AF/FB = p/q, and BD/DC = q/r, and CE/EA = r/p.
Hence, multiplying, we see that the formula is true numerically.
Conversely, if three points D, E, F, taken on the sides BC, CA, AB of a triangle,
satisfy the relation
AF . BD . CE = – FB . DC . EA,
then D, E, F are collinear.
For, if not, let DE cut AB in F ′. Then since D, E, F ′ are collinear, we have
AF ′ . BD . CE = – F ′B . DC . EA.
But by hypothesis we have
AF . BD . CE = – FB . DC . EA.
Dividing, we get AF ′ : F ′B : : AF : FB; hence
AF ′ + F ′B : AF + FB :: AF ′ : AF,
i.e.
AF ′ = AF, i.e. F ′ coincides with F. Hence D, E, F are collinear.
6
AN ELEMENTARY TREATISE ON PURE GEOMETRY WITH NUMEROUS EXAMPLES
EXAMPLE 1. Show that the above relation is equivalent to
sin ACF . sin BAD . sin CBE
= – sin FCB . sin DAC . sin EBA.
For
AF : FB = Δ ACF : Δ FCB
1
1
= AC. CF sin ACF : FC. CB sin FCB.
2
2
EXAMPLE 2. If any transversal cut the sides AB, BC, CD, DE, ... of any polygon in
the points a, b, c, d, ..., show that the continued product of the ratios
Aa/Ba, Bb/Cb, Cc/Dc, Dd/Ed, ...is unity.
Let AC cut the transversal in γ, AD in δ, and so on,
then
Aa/Ba × Bb/Cb × Cγ/Aγ = 1
and Aγ/Cγ × Cc/Dc × Dδ/Aδ = 1, and so on.
Multiplying up and cancelling, we get the theorem.
EXAMPLE 3. If on the four lines AB, BC, CD, DA there be taken four points a, b, c,
d such that
Aa . Bb . Cc . Dd = aB . bC . cD . dA,
show that ab and cd meet on AC and ad and bc meet on BD.
Apply Menelaus’s Theorem to ABD and ad and to BCD and bc; multiply, and divide
by the given relation; and we see that ad and bc meet BD in the same point; similarly
for AC.
EXAMPLE 4. If the sides of the triangle ABC which is inscribed in a circle be cut by
any transversal in D, E, F, show that the product of the tangents from D, E, F to the
circle is numerically equal to AF . BD . CE.
EXAMPLE 5. Construct geometrically the ratio a/b ÷ c/d.
Divide AB in the ratio a/b and AC in the ratio c/d.
EXAMPLE 6. The bisectors of two angles of a triangle and the bisector of the
supplement of the third angle meet the opposite sides in collinear points.
CEVA’S THEOREM
P
7. If the lines joining any point to the vertices A, B, C of a triangle meet the
opposite sides in D, E, F, then
AF . BD . CE = FB . DC . EA.
To verify the sign of the formula. O the point in which AD, BE, CF meet must be
either inside the triangle, in which case each of the ratios AF : FB and BD : DC and
CE : EA is positive, or as at O1 or O2, in which cases two of the ratios are negative
and one positive. Hence the sign of the formula is correct.
To prove the formula numerically, we have
AF : FB : : ΔACF : ΔFCB : : ΔAOF : ΔFOB
: : ΔACF – ΔAOF : ΔFCB – ΔFOB
: : ΔAOC : ΔBOC.
7
PURE GEOMETRY
F
A
O1
E
F
E
O
O2
Similarly
and
B
D
C
D
BD : DC : : Δ BOA : Δ AOC
CE : EA : : Δ COB : Δ AOB.
Hence, multiplying, we see that the formula is true numerically.
Conversely, if three points D, E, F, taken on the sides BC, CA, AB of a triangle,
satisfy the relation
AF . BD . CE = FB . DC . EA,
then AD, BE, CF are concurrent.
For, if not, let AD, BE cut in O; and let CO cut AB in F ′. Then since AD, BE,
CF ′ are concurrent, we have
AF′ . BD . CE = F′B . DC. EA.
But by hypothesis we have
AF . BD . CE = FD . DC . EA.
Dividing, we get AF′ : F′B : : AF : FB. Hence F and F ′ coincide, i.e. AD, BE, CF
are concurrent.
Example 1. In the figure, show that
OD OE OF
+
+
= 1.
AD BE CF
Example 2. Show that the necessary and sufficient condition that Aa, Bb, Cc should
meet in a point is sin aAB . sin bBC . sin cCA = sin CAa . sin ABb . sin BCc.
Example 3. If the lines Aa, Bb, Cc, Dd, ... drawn through the vertices of a plane
polygon ABCD ... in the same plane meet in a point, then the continued product of
such ratios as sin aAB : sin ABb is unity.
Example 4. If the lines joining a fixed point O to the opposite vertices of a polygon
of an odd number of sides meet the sides AB, BC, CD, DE, ... in the points a, b, c,
d,...., show that the continued product of such ratios as Aa/aB is unity.
For Aa/aB = AO . aO sin AOa/aO . BO sin aOB.
Example 5. A circle meets BC in D, D′, CA in E, E′ and AB in F, F ′. Show that if
AD, BE, CF meet in a point, so do AD′, BE′, CF ′.
8
AN ELEMENTARY TREATISE ON PURE GEOMETRY WITH NUMEROUS EXAMPLES
THE STRAIGHT LINE AT INFINITY
P
8. Take any fixed point O and a fixed line l. Then any line x through O cuts l in
a point P. Now rotate x about O so that x may become more and more nearly parallel
to l. Then P recedes indefinitely along l; and in the limit when x is parallel to l, P is
said to be the point at infinity upon l. Hence two parallel lines intersect in a point at
infinity. Notice that the two points at infinity on a line are coincident; for instance in
the figure each point is the limit of P when x is parallel to l.
O
x
l
P
Now draw a plane β parallel to any plane α. This will be parallel to every line in
the plane α. Hence the point at infinity on each line in α lies on the plane β. Now
every two planes intersect in a line. Hence all the points at infinity in the plane a lie
on a single straight line (called the straight line at infinity in this plane), viz. the line
of intersection of the planes α and β,
P 9. The point at infinity on the line of any segment divides this segment externally
in the ratio – 1.
A
B
P
For let P divide the segment AB externally. Then
AP
AP
AB + BP
AB
=–
=–
=–
– 1.
PB
BP
BP
BP
Also when P is at infinity, the limit of AB/BP is zero; for AB is finite and BP is
infinite.
We may say that a segment is bisected externally by the point at infinity on its
line.
EXERCISE 1
1. If A, B, C are the angles of a triangle and A′, B′, C′ the angles which the sides BC, CA,
AB make with any line, then sin A . sin A′ + sin B . sin B′ + sin C . sin C′ = 0.
2. OL, OM, ON are any three lines through O and PL, PM, PN make equal angles with
OL, OM, ON in the same way; show that
PL . sin MON + PM . sin NOL + PN . sin LOM = 0.
3. If A, B, C are the angles of a triangle and A′, B′, C′ the angles which the sides BC, CA,
AB make with any line,
then
sin A′ . sin B ′
sin C ′ . sin A′
sin B ′ . sin C ′
+
+
= – 1.
sin A . sin B
sin C . sin A
sin B . sin C
PURE GEOMETRY
9
4. If OA, OB, OC are any three lines through O, and PA, PB, PC are the three perpendiculars from any point P to OA, OB, OC, then
• Σ {PB . PC sin BOC} = – PO2 sin BOC . sin COA . sin AOB.
5. Through the vertices A, B, C of a triangle are drawn the parallels AX, BY, CZ to meet the
sides BC, CA, AB at X, Y, Z; show that
BX . CX
+
CY . AY
+
AZ . BZ
= 1.
AX
BY
CZ 2
6. If VA, VB, VC, VD, VO, VO′ are any six concurrent lines, and if α = sin OVA ÷ sin O′VA,
and so on, show that
2
2
sin AVC sin BVC γ – α γ – β
.
÷
=
÷
sin AVD sin BVD δ – α δ – β
7. Three lines OAA′, OBB′, OCC′ are cut by two lines ABC, A′B′C′; show that
OA AB OA′ A′ B′
÷
=
÷
OC BC OC ′ B′ C ′
AA′ . BC BB′ . CA CC ′ . AB
+
+
= 0.
OA′
OB′
OC ′
If the polygon A′B′C′D′... is inscribed in the polygon ABCD ..., so that A′ is on AB, B′
on BC, and so on, and O is any point in the plane, show that the continued product of
such ratios as sin AOA′/ sin A′OB ÷ AA′/A′B is unity.
If n points A, B, C, ... and n′ points A′, B′, C′, ... are collinear, and if G and G′ are points
on the line AB ... such that GA + GB + GC + ... = 0, and G′ A′ + G′B′ + G′C′ + ... = 0,
show that n . n′. GG′ = Σ (AA′ + AB′ + AC′ + ...).
If A, B, X, Y are collinear points such that AX . AY = BX . BY, then AB and XY have the
same bisector unless A and B or X and Y coincide.
A line, meets BC, CA, AB at D, E, F, P, Q, R bisect EF, FD, DE. AP, BQ, CR meet BC,
CA, AB at X, Y, Z. Show that X, Y, Z are collinear.
A transversal meets the sides of a polygon ABCD ... at A′, B′, C′, ... and meets any lines
through the vertices A, B, C, ... at A′′, B′′, C′′, ... ; show that the continued product of
such ratios as sin A′BB′′/sin B′′BB′ ÷ A′B′′/B′′B′ is unity.
If the lines AB, BC, CD, DA, which are not in the same plane, be met by any plane at A′,
B′, C′, D′, then AA′. BB′. CC′. DD′ = A′B. B′C. C′D. D′A.
AO meets BC at D, BO meets CA at E, CO meets AB at F. GH is equal and parallel to
BC and passes through A. BC meets GO at L and HO at K. Similarly segments such as
KL are constructed on CA and AB. Show that the product of these segments is
AF. BD. CE.
AO meets BC at P, BO meets CA at Q, CO meets AB at R. PU meets QR at X, QU meets
RP at Y, RU meets PQ at Z. Show that AX, BY, CZ are concurrent.
Through the vertices of a triangle are drawn parallels to the reflexions of the opposite
sides in any line. Show that these parallels are concurrent.
On the sides AB, AC of a triangle are taken the points X, Y, and in BC a point P is taken
such that AB. AY. PC = AC. AX. BP. Show that AP will bisect XP.
and
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
CHAPTER
2
Harmonic Ranges and Pencils
P
1. A range or row is a set of points on the same line, called the axis or base of
the range.
A pencil is a set of lines, called rays, passing through the same point, called the
vertex or centre of the pencil.
If A, B, A′, B′ are collinear points such that
AB : BA′ : : AB′ : A′B′
or (which is the same thing) such that
AB/BA′ = – AB′ / B′A′,
then (ABA′B′) is called a harmonic range. A, A′ and B, B′ are called harmonic pairs
of points; and A, A′ are said to be conjugate, as also B, B′. Also A is said to be the
fourth harmonic of A′ (and A′ of A) for B and B′; so B is said to be the fourth harmonic
of B′ (and B′ of B) for A and A′. Also AA′ and BB′ are called harmonic segments and
are said to divide one another harmonically. The briefest and clearest way of stating
the harmonic relation is to say that (AA′, BB′) is harmonic. The relation may be stated
in words thus — each pair of harmonic points divides the segment joining the other
pair in the same ratio internally and externally.
For
A
B
A¢
BA : AB¢ = –BA¢ : A¢B¢
B¢
EXAMPLE 1. The centres of similitude of two circles divide the segment joining the
centres of the circles harmonically.
EXAMPLE 2. (BC, XX ′), (CA, YY ′), (AB, ZZ′) are harmonic ranges; show that if
AX, BY, CZ are concurrent, then X ′Y′Z′ are collinear, and that if X ′Y ′Z′ are collinear,
then AX, BY, CZ are concurrent.
Use the theorems of Ceva and Menelaus.
P 2. If (AA′, BB′) be harmonic, then
2
1
1
1
1
2
=
+
+
=
,
AB AB ′ BB′ BA BA′
AA′
2
2
1
1
1
1
=
+
+
=
,
A′ A
A ′ B A ′ B ′ B′ B B′ A B′ A′
10