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There are 10 theorems
to be proved on our
course:
1. verticary opposite angres
are equar in measure.
The measu res of the
three angles of a triangle
sum to 1g0o.
3. An exteriot r angle of a triangle
equals the sum of the
measure.
two interior
2.
4.
If two sides
equal
5'
6.
7'
in;J.f"*tngle
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A diagonal
opposite angres in
are'equal in measure'
then the angres opposite
these sides are
sides and opposite
angles of a paralrerogram
are respectively equar
in
bisects the area of
a parallelogram.
;:'"ilffffiJ,1i::#f1",n: ff;:T.#the.circre
is rwice the measure
orthe angre ar
There are three deductions
associated with this
theorem:
Deduction 1:
AII angres at the ciicumfere
ce on the same afc
are equar in measure.
Deduction 2:
An angle subtended by
a diame Er at thecircumference
is a right angle.
Deduction 3:
The sum of opposite
angles of a cycric quadrilateral
is 1g0..
A fine through the centre
of u
pelpendicurar to a
chord bisects the chord.
If two triangles are equiangular,"i."1.
the lengths of corresponding
sides are in proportion.
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Theorem:
Vertically opposite angles are equal in measure.
Given:
To prove:
Construction:
Proof:
Intersecting lines
lttl: tzl
L
and
K, with vertically opposite angles
I
and 2.
I
Label angle 3.
tIl+ll3l=1gg'
t2l
+ 1231: 1gg
ttl+lz3l=ltzl+lt3l
tll: t2l
straight angle
straight angle
I
1
The measures of the three angles of a triangle sum to 180'.
Given:
To prove:
Construction:
Proof:
Labc with angles 1,2 and3.
l ltl + l t2l + 1t31= 199".
Draw a line througlt a, parallel to bc. Label angles 4 and 5.
alternate angles
and I t2l = | Z5l
| 1 tl = |
:
+
1
+
t3l
+
131
tal
sl
tl
+
z2l
|
|
I
|
I
I
straight angle
141
+
/.31
199".
+
t5l
=
|
|
|
t
t4l
lztl+lt2l+lt3l=199".
Theorem:
An exterior angle of a triangle equals the sum of the two interior
opposite angles in measure.
Given:
To prove:
Construction:
Proof:
Aabc with interior opposite angles
lttl+lZ2l:lt3l
I
and 2 and exterior angle 3.
Label angle 4.
lztl + l2l + 1241= 139.
lz3l + l4l = 1gg
lltl+ t2l+lt4l=lt3l+lZ4l
lzll
Theorem:
+
three angles in a triangle
straight angle
,/)l=L/?l
If two sides of a triangle are equal in measure, then the angles
opposite these sides are equal in measure.
Given:
To prove:
Construction:
Proof:
Aabc, with labl : lctcl and base angles I
and 2.
lttl=l/21
Draw ad, the bisector of /.bac. Label angles 3 and 4.
Consider Labd and Lacd:
glven
labl = lacl
construction
lt3l:1141
common
ladl: ladl
Aabd = Lacd
I
ztl
= 1121
SAS
corresponding angles
Opposite sides and opposite angles of a parallelogram are
respectively equal in measure.
I
4
3
Given:
To prove:
2
Parallelogram abcd
labl=ldcl, ladl:lbcl
Z abcl = | I adcl, I Z badl : I Z bcdl
Join a to c. Label angles 1,2,3 and4.
Consider Aabc and Ladc:
and I Z3l I z4l
I Zll = |
I
Construction:
Proof:
:
t2l
lacl: lacl
Aabc: Aadc
ASA
labl = ldcl and
corresponding sides
corresponding angles
andlZabcl=lladcl
similarly, I Z badl
:
alternate angles
common
|
1 bcdl
A diagonal bisects the area of a parallelogram.
Parallelogram abcd with diagonal lacl.
Area of Aabc =Area of Aadc.
Consider Labc and Aadc:
opposite sides
labl = ldcl
opposite sides
ladl = lbcl
common
lacl = lacl
Aabc: Aadc
Area Aabc =Area Aadc.
SSS
] The measure
Ff
Given:
To prove:
Construction:
Proof:
,h"
""gtq
of the angle at the centre of the circle is twice the measure
,h" .k"mference, standing on the same arc.
"
Circle, centre o, containing points a, b and c.
I
lbocl:2l lbacl
Join a to o and continue to d. Label angles l, 2, 3, 4 and 5.
Consider Aaob:
exterior angle
ltll=lz2l+lt3l
but lz.2l =
|
t3l
IZII=2lt2l
similarly, l14l=2ll5l
ltll + z4l:zl t2l +21z5l
t
I
= 2(l z2l + | tsl)
le I lbocl:21lbacl
I
tl + | z4l
loal:
lobl
All
angles at the circumference on the same arc are equal in measure.
To prove:
Proof:
I
lbacl
:
I
lbdcl
lt3l =2lztl
I
t al
ILJI -)l
,/
)l
2lttl :21 l2l
I
trl :l,/)l
|
angle at the centre is twice the angle
on the circumference (both on arc bc)
angle at the centre is twice the angle
on the circumference (both on arc bc)
-l
i.e. llbacl : lbdcl
I
Deduction
An angle subtended by a diameter at the circumference is a right angle.
To prove:
Proof:
lZbacl =)Q"
lt2l =2lZrl
but
i.e.
I
tZl
I
zrl
2ltrl
I
lbacl
= 180o
= 180o
angle at the centre is twice the angle on
the circumference (both on arc bc)
straight angle
= 90o
= 90o
l
Deduction 3:
The sum of the opposite angles of a cyclic quadrilateral is 180".
L
To prove:
llbadl+ltbcdl:
Proof:
lt3l =2lztl
angle at the centre is twice the
z4l =21Z2l
angle on the circumference
(both on minor arc bd)
angle at the cenffe is twice the
angle on the circumference
(both on major arc bd)
I
t
z.3l + | z4l = 2l Il + 2l tzl
angles at a point
I l3l + | 141 = 369"
I
but
2l
i.e.
'
180o
t Il + 21 t2l :360
ltll+ll2l =1gg'
lZbadl
+llbcdl=
180"
Similarly, lZabcl + l Z.adcl=
180o
A line through the centre of
a circle perpendicular to a chord bisects
the chord.
a
1__l
Given:
d
Circle, centre c, a line 1- containing c, chord fabf,
such that L Lab and Loab: d.
To prove:
ladl:
Construction:
Proof:
Label right angles 1 and 2.
Consider Lcda and Acdb:
I
lbdl
Z1l: 221:99"
lcbl
lcdl
glven
both radii
common
Acdb
RHS
lbdl
corresponding sides
)
lcal:
)cdl:
Acda
:
)adl:
Theorem:
If two triangles
are equiangular, the lengths of corresponding sides are in
proportion.
b
Given:
Equiangular triangles abc and xyz in which
=I
I
I t2l
| L5l and I t3l = | t6l.
ttl
:
t4|
_lbcl
lv'l
Construction:
Mark the point m on labl such that laml = lxyl.
Mark the point n on lacl such that lanl = lxzl.
Jornmto n.Label angle 7.
Consider
Layz
and
Lxyz:
loyl : lxyl and lazl :
lztl=lz4l
lxzl
construction
given
Layz= Lxyz
SAS
ltTl=lzsl
corresponding angles
glven
t,/)t:t,/\l
|--|l-.I
ltll=lt2l
yz ll bc
lobl
locl
lo*l
lorl
l"bl
locl
lrryl lrrl
A line parallel to one side
divides the other two sides
in the same proportion
laml: l-ryl and lanl:lxzl
In
of the length of the side opposite to
the right angle is equal to the sum of the squares of the lengths of the other
two sides.
Given:
To prove:
Construction:
right-angled triangle,
a right-angled
le, the square
Righrangled triangle with length of sides a, b and c as shown.
a2+b2:cz
Draw a square with sides of length a + b.
Draw four congruent right angled triangles in the square
with sides of length a and b and hypotenuse c, as shown.
Label angles 1,2,3 and 4.
lll + 1221=99"
lztl=lt3l
remaining angles
coresponding angles
l
I
z2l + | Z3l =99'
Area of square
=
=
\ote: A difficulty with the proof
:
b)'
= 4(area
of one triangle) + c2
(a + b)2
:41+ab
1+ c2
(a +
a2+2ab+b2:2ab+c2
is trying to draw the diagram. One way to do this is to
let a=2cm, b =5cm and draw a square with each side 7cm in length. Then
simply mark off 2 cm on each side in clockwise direction. Join these points to
construct the smaller square.