AoPS Introduction to Geometry. Chapter 3 Congruent Triangles

AoPSIntroduction 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

<B ≅ <E

<C ≅ <F

AB ≅ DE

BC ≅ EF

CA ≅ FD

ΔABC ≅ ΔDEF

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

<B ≅ <E

<C ≅ <F

AB ≅ DE

BC ≅ EF

CA ≅ FD

ΔABC ≅ ΔDEF

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

17

99

25°

CPCTC


Find <CDB

17

17

99

25°

Since DA = CB,AB = CD, andDB = DB, we haveΔABD ≅ ΔCDB bySSS.

CPCTC


Find <CDB

17

17

99

25°


Since <CDB = <ABD,are correspondingparts of these 2 Δ’s,they must be equal.

Therefore, <CDB = <ABD = 25°

CPCTC


Find <CDB

17

17

99

25°


Since <CDB = <ABD,are correspondingparts 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

555° 55°35° 35°

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

Δ ACM ≅ Δ ABM. A

C

5 5

B

30°

Isosceles Triangles



Δ ACM ≅ Δ ABM. A

C

5 5

B

30°

M

Isosceles Triangles



Δ ACM ≅ Δ ABM.

AB ≅ AC

BM ≅ CM

AM ≅ AM

so Δ ACM ≅ Δ ABM

A

C

5 5

B

30°

M

Isosceles Triangles



Δ ACM ≅ Δ ABM.

AB ≅ AC

BM ≅ CM

AM ≅ AM

so Δ ACM ≅ Δ ABM

A

C

5 5

B

30°

M

by CPCTC, < B ≅ < C

Isosceles Triangles



Δ ACM ≅ Δ ABM.

AB ≅ AC BC is called the

BM ≅ CM base of the isosceles

AM ≅ AM triangle and

so Δ ACM ≅ Δ ABM < B & < C are known

as base angles.

A

C

5 5

B

30°

M

by CPCTC, < B ≅ < C

Problem 3.17

AC = CD = DB, and < B = 23°. Find < A.

A

C B

D

23°

Problem 3.17


A

C B

D

23°

Since DC = BC, we have< DCB = < B = 23°. We couldfind < CDB to get < ADC, orwe can note that < ADC is anexterior angle of Δ BCD, so

Problem 3.17


< ADC = < DCB + < B = 46°. Since AC = DC, we have

< A = < ADC = 46°.

A

C B

D

23°

Since DC = BC, we have< DCB = < B = 23°. We couldfind < CDB to get < ADC, orwe can note that < ADC is anexterior angle of Δ BCD, so

Problem 3.18

XY = XZ = 8 and < X = 60°. Find YZ.

Y Z

X

8 8

60°

Problem 3.18

Since <X = 60°, <Y + <Z = 180° - 60° = 120°

Since XY = XZ, then <Y = <Z, so <Y = <Z = 60°.

Y Z

X

8 8

60°

Problem 3.18

Therefore, all angles of Δ XYZ are equal.

Specifically, <X = <Y means YZ = XZ, so YZ = 8.Y Z

X

8 8

60°

Important!

If all three angles of a triangle are equal, then so

are all 3 sides.

Conversely, if all three sides are equal, then all

three angles equal 60°.

60°

60°60°

Problem 3.19

Let’s put our isosceles and equilateral triangles to

work. Find <PBD.P

BA

D C

ΔPAB is an equilateral triangle andABCD is a square.

Problem 3.19

Let’s put our isosceles and equilateral triangles to

work. Find <PBD.P

BA

D C


Draw BD so we can see <PBD.

Problem 3.19

Since <BAD = 90°, the other 2 angles ΔBAD =

(180 – 90)/2 = 45°.

P

BA

D C


Draw BD so we can see <PBD.Since ΔPAB is equilateral, <PBA =60°.ΔBAD is isosceles, since AB = AD.

Problem 3.19 Therefore: <PBD = <PBA + <ABD = 105°

Since <BAD = 90°, the other 2 angles ΔBAD =

(180 – 90)/2 = 45°.

P

BA

D C


Draw BD so we can see <PBD.Since ΔPAB is equilateral, <PBA =60°.ΔBAD is isosceles, since AB = AD.

Problem 3.20O is the center of the circle, AB || CD, <ABO = 24°, and

<OBC = <OCD. Find <BOC.

24°



24°

We have parallel lines and an isoscelestriangle (OB = OC because they areradii of the same circle). So we can dosome angle-chasing. We let <OBC = xso that we also have <OCD = x fromthe given information and <OCB = x from isosceles ΔOCB.



24°


Since AB || CD, we have <ABC + <BCD = 180°. Therefore, <ABO + <OBC + <BCO + <OCD = 180°.



24°


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


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




(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




(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) = 180°. So x = 28⅖°.

Challenge 3.44ΔABC 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.




So < ACD = 90° - (<CAB)/2. We also have <CAB = 180 - <ACB - <ABC = 90° - <B, so <ACD = <ADC = 45° + (<B/2).




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).




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<ACE = 90 - <BCE = <B/2.

Challenge 3.44Therefore, <DCE = 90 - <ACE - <BCD =

90 - (<B/2) – (45 - (<B/2) ) = 45°.


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<ACE = 90 - <BCE = <B/2.

Fini! The end of Chapter 3!

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AoPS Introduction to Geometry. Chapter 3 Congruent Triangles