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AoPS
Introduction to
Geometry
Chapter 3
Congruent Triangles
Congruent
Two figures are congruent if they are exactly the same:
we can slide, spin, and/or flip so that it is exactly on top
of the other figure.
Congruent Triangles
If we have two congruent triangles, all the
corresponding pairs of sides are equal, as are the
corresponding angles.
<A ≅ <D
ΔABC ≅ ΔDEF
AB ≅ DE
<B ≅ <E
BC ≅ EF
<C ≅ <F
CA ≅ FD
Congruent Triangles
Conversely, if all pairs of corresponding sides of 2
triangles have equal lengths, and the corresponding
angles of the 2 triangles are equal, then the triangles
are congruent.
<A ≅ <D
ΔABC ≅ ΔDEF
AB ≅ DE
<B ≅ <E
BC ≅ EF
<C ≅ <F
CA ≅ FD
SSS Congruence
If three sides of one triangle are congruent to the
corresponding sides of a second triangle, then the triangles
are congruent.
AB ≅ A’B’
BC ≅ B’C’
CA ≅ C’A’
ΔABC ≅ ΔA’B’C’ by SSS
CPCTC
In the figure, the sides and angle have measures shown.
Find <CDB.
17
9
9
25°
17
CPCTC
In the figure, the sides and angle have measures shown.
Find <CDB
17
Since DA = CB,
AB = CD, and
DB = DB, we have
ΔABD ≅ ΔCDB by
SSS.
9
9
25°
17
CPCTC
In the figure, the sides and angle have measures shown.
Find <CDB
17
Since DA = CB,
AB = CD, and
DB = DB, we have
ΔABD ≅ ΔCDB by
SSS.
9
9
25°
17
Since <CDB = <ABD,
are corresponding
parts of these 2 Δ’s,
they must be equal.
Therefore,
<CDB = <ABD = 25°
CPCTC
In the figure, the sides and angle have measures shown.
Find <CDB
17
Since DA = CB,
AB = CD, and
DB = DB, we have
ΔABD ≅ ΔCDB by
SSS.
9
9
25°
17
Since <CDB = <ABD,
are corresponding
parts of these 2 Δ’s,
they must be equal.
Therefore,
<CDB = <ABD = 25°
Notice that we can now use <CDB = <ABD to prove that CD || BA.
CPCTC
The previous problem shows us a very important general
use of congruent triangles. Once we determine that 2
triangles are congruent (≅), we can apply whatever we
know about the sides and angles of one triangle to the
other triangle. This obvious principle goes by the name
Corresponding Parts of Congruent Triangles are Congruent,
or CPCTC.
SAS Congruence
If 2 sides and the included angle of one triangle are
congruent to the corresponding 2 sides and included angle
of another triangle, then the triangles are congruent.
AC ≅ A’C’
<C ≅ <C’
BC ≅ B’C’
So ΔABC ≅ ΔA’B’C’ by SAS.
ASA Congruence
If 2 angles and the included side of one triangle are
congruent to the corresponding 2 angles and included side
of another triangle, then the triangles are congruent.
<B ≅ <B’
BC ≅ B’C’
<C ≅ <C’
ΔABC ≅ ΔA’B’C’ by ASA
AAS Congruence
If 2 angles and a non-included side of one triangle are
congruent to the corresponding 2 angles and non-included
side of another triangle, then the triangles are congruent.
<A ≅ <A’
<C ≅ <C’
AB ≅ A’B’
ΔABC ≅ ΔA’B’C’ by AAS
Not AAS Congruence
The triangles shown are NOT congruent because the
equal sides are NOT adjacent to corresponding equal
angles!
5
55°
35°
55° 35°
5
Angle Side Side Congruence
There is NO Congruence theorem known as
Angle Side Side (or SSA)
An easy way to remember this is to think
there are no donkeys in geometry!
Isosceles Triangles
In the diagram, Δ XYZ is an isosceles triangle.
XY ≅ XZ
Isosceles Triangles
In the diagram, AB = AC = 5 and <CAB = 30°.
Let M be the midpoint of BC. Draw AM. Show that
A
Δ ACM ≅ Δ ABM.
30°
5
B
5
C
Isosceles Triangles
In the diagram, AB = AC = 5 and <CAB = 30°.
Let M be the midpoint of BC. Draw AM. Show that
A
Δ ACM ≅ Δ ABM.
30°
5
B
5
M
C
Isosceles Triangles
In the diagram, AB = AC = 5 and <CAB = 30°.
Let M be the midpoint of BC. Draw AM. Show that
A
Δ ACM ≅ Δ ABM.
30°
AB ≅ AC
BM ≅ CM
AM ≅ AM
so Δ ACM ≅ Δ ABM
5
B
5
M
C
Isosceles Triangles
In the diagram, AB = AC = 5 and <CAB = 30°.
Let M be the midpoint of BC. Draw AM. Show that
A
Δ ACM ≅ Δ ABM.
30°
AB ≅ AC
BM ≅ CM
AM ≅ AM
so Δ ACM ≅ Δ ABM
5
B
5
M
C
by CPCTC, < B ≅ < C
Isosceles Triangles
In the diagram, AB = AC = 5 and <CAB = 30°.
Let M be the midpoint of BC. Draw AM. Show that
A
Δ ACM ≅ Δ ABM.
30°
AB ≅ AC
BM ≅ CM
AM ≅ AM
so Δ ACM ≅ Δ ABM
5
B
M
BC is called the
5 base of the isosceles
triangle and
< B & < C are known
C as base angles.
by CPCTC, < B ≅ < C
Problem 3.17
AC = CD = DB, and < B = 23°. Find < A.
A
D
C
23°
B
Problem 3.17
AC = CD = DB, and < B = 23°. Find < A.
A
D
C
23°
B
Since DC = BC, we have
< DCB = < B = 23°. We could
find < CDB to get < ADC, or
we can note that < ADC is an
exterior angle of Δ BCD, so
Problem 3.17
AC = CD = DB, and < B = 23°. Find < A.
A
D
C
23°
B
Since DC = BC, we have
< DCB = < B = 23°. We could
find < CDB to get < ADC, or
we can note that < ADC is an
exterior angle of Δ BCD, so
< ADC = < DCB + < B = 46°. Since AC = DC, we
have
< A = < ADC = 46°.
Problem 3.18
XY = XZ = 8 and < X = 60°. Find YZ.
X
60°
8
Y
8
Z
Problem 3.18
Since <X = 60°, <Y + <Z = 180° - 60° =
X
120°
60°
8
Y
8
Z
Since XY = XZ, then <Y = <Z, so <Y = <Z = 60°.
Problem 3.18
Therefore, all angles of Δ XYZ are equal.
X
60°
8
Y
8
Z
Specifically, <X = <Y means YZ = XZ, so YZ = 8.
Important!
If all three angles of a triangle are equal, then so
are all 3 sides.
60°
60°
60°
Conversely, if all three sides are equal, then all
three angles equal 60°.
Problem 3.19
Let’s put our isosceles and equilateral triangles
to
work. Find
<PBD.
ΔPAB is an equilateral triangle and
P
ABCD is a square.
A
B
D
C
Problem 3.19
Let’s put our isosceles and equilateral triangles
to
work. Find
<PBD.
ΔPAB is an equilateral triangle and
P
ABCD is a square.
A
B
Draw BD so we can see <PBD.
D
C
Problem 3.19
ΔPAB is an equilateral triangle and
ABCD is a square.
P
A
D
B
C
Draw BD so we can see <PBD.
Since ΔPAB is equilateral, <PBA =
60°.
ΔBAD is isosceles, since AB = AD.
Since <BAD = 90°, the other 2 angles ΔBAD =
(180 – 90)/2 = 45°.
Problem 3.19
Therefore: <PBD = <PBA + <ABD = 105°
ΔPAB is an equilateral triangle and
ABCD is a square.
P
A
D
B
C
Draw BD so we can see <PBD.
Since ΔPAB is equilateral, <PBA =
60°.
ΔBAD is isosceles, since AB = AD.
Since <BAD = 90°, the other 2 angles ΔBAD =
(180 – 90)/2 = 45°.
Problem 3.20
O is the center of the circle, AB || CD, <ABO = 24°, and
<OBC = <OCD. Find <BOC.
24°
Problem 3.20
O is the center of the circle, AB || CD, <ABO = 24°, and
<OBC = <OCD. Find <BOC.
24°
We have parallel lines and an isosceles
triangle (OB = OC because they are
radii of the same circle). So we can do
some angle-chasing. We let <OBC = x
so that we also have <OCD = x from
the given information and <OCB = x
from isosceles ΔOCB.
Problem 3.20
O is the center of the circle, AB || CD, <ABO = 24°, and
<OBC = <OCD. Find <BOC.
24°
We have parallel lines and an isosceles
triangle (OB = OC because they are
radii of the same circle). So we can do
some angle-chasing. We let <OBC = x
so that we also have <OCD = x from
the given information and <OCB = x
from isosceles ΔOCB.
Since AB || CD, we have <ABC + <BCD = 180°. Therefore,
<ABO + <OBC + <BCO + <OCD = 180°.
Problem 3.20
O is the center of the circle, AB || CD, <ABO = 24°, and
<OBC = <OCD. Find <BOC.
24°
We have parallel lines and an isosceles
triangle (OB = OC because they are
radii of the same circle). So we can do
some angle-chasing. We let <OBC = x
so that we also have <OCD = x from
the given information and <OCB = x
from isosceles ΔOCB.
Since AB || CD, we have <ABC + <BCD = 180°. Therefore,
<ABO + <OBC + <BCO + <OCD = 180°.
Substitution gives 24° + 3x = 180°, so x = 52°.
So <BOC = 180° - 2x = 76°.
Challenge 3.41
The measures of two angles of an isosceles triangle
are 3x + 4° and x + 17°. Find all possible values of
x.
Challenge 3.41
The measures of two angles of an isosceles triangle
are 3x + 4° and x + 17°. Find all possible values of
x.
There are 3 possible cases:
(a) The 2 angles given are equal. In this case, we have 3x + 4
= x + 17, so x = 6½°.
Challenge 3.41
The measures of two angles of an isosceles triangle
are 3x + 4° and x + 17°. Find all possible values of
x.
There are 3 possible cases:
(a) The 2 angles given are equal. In this case, we have 3x + 4
= x + 17, so x = 6½°.
(b) There are 2 angles with measure 3x + 4°. The sum of
the 3 angles of the Δ must be 180°, so we have 2(3x + 4)
+ x + 17 = 180. Therefore x = 28⅖
Challenge 3.41
The measures of two angles of an isosceles triangle
are 3x + 4° and x + 17°. Find all possible values of
x.
There are 3 possible cases:
(a) The 2 angles given are equal. In this case, we have 3x + 4
= x + 17, so x = 6½°.
(b) There are 2 angles with measure 3x + 4°. The sum of
the 3 angles of the Δ must be 180°, so we have 2(3x + 4)
+ x + 17 = 180. Therefore x = 22 1/7°.
(c) There are 2 angles with measure x + 17°. The sum of
the angles must be 180, so we have 3x + 4 + 2(x + 17) =
Challenge 3.44
ΔABC has a right angle at <C. Points D and E are on
AB such that AD = AC and BE = BC. Find <DCE.
Challenge 3.44
ΔABC has a right angle at <C. Points D and E are on
AB such that AD = AC and BE = BC. Find <DCE.
Since AD = AC, <ACD = <ADC.
Also <ACD + <ADC + <CAD =
180°, so 2<ACD = 180 - <CAD =
180 - <CAB.
So < ACD = 90° - (<CAB)/2.
Challenge 3.44
ΔABC has a right angle at <C. Points D and E are on
AB such that AD = AC and BE = BC. Find <DCE.
Since AD = AC, <ACD = <ADC.
Also <ACD + <ADC + <CAD =
180°, so 2<ACD = 180 - <CAD =
180 - <CAB.
So < ACD = 90° - (<CAB)/2. We also have
<CAB = 180 - <ACB - <ABC = 90° - <B, so
<ACD = <ADC = 45° + (<B/2).
Challenge 3.44
ΔABC has a right angle at <C. Points D and E are on
AB such that AD = AC and BE = BC. Find <DCE.
Since AD = AC, <ACD = <ADC.
Also <ACD + <ADC + <CAD =
180°, so 2<ACD = 180 - <CAD =
180 - <CAB.
So < ACD = 90° - (<CAB)/2. We also have
<CAB = 180 - <ACB - <ABC = 90° - <B, so
<ACD = <ADC = 45° + (<B/2). Then, <BCD = <BCA <ACD = 90° - <ACD = 45° - (<B/2).
Challenge 3.44
ΔABC has a right angle at <C. Points D and E are on
AB such that AD = AC and BE = BC. Find <DCE.
Since AD = AC, <ACD = <ADC.
Also <ACD + <ADC + <CAD =
180°, so 2<ACD = 180 - <CAD =
180 - <CAB.
So < ACD = 90° - (<CAB)/2. We also have
<CAB = 180 - <ACB - <ABC = 90° - <B, so
<ACD = <ADC = 45° + (<B/2). Then, <BCD = <BCA <ACD = 90° - <ACD = 45° - (<B/2). From isosceles
ΔBCE, we have <BCE = <BEC = (180 - <B)/2 = 90 - (<B/2),
so
Challenge 3.44
Therefore, <DCE = 90 - <ACE - <BCD =
90 - (<B/2) – (45 - (<B/2) ) = 45°.
Since AD = AC, <ACD = <ADC.
Also <ACD + <ADC + <CAD =
180°, so 2<ACD = 180 - <CAD =
180 - <CAB.
So < ACD = 90° - (<CAB)/2. We also have
<CAB = 180 - <ACB - <ABC = 90° - <B, so
<ACD = <ADC = 45° + (<B/2). Then, <BCD = <BCA <ACD = 90° - <ACD = 45° - (<B/2). From isosceles
ΔBCE, we have <BCE = <BEC = (180 - <B)/2 = 90 - (<B/2),
so
Fini! The end of Chapter 3!