Introduction to Geometry 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. ΔABC ≅ ΔDEF <A ≅ <D <B ≅ <E <C ≅ <F AB ≅ DE BC ≅ EF 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. ΔABC ≅ ΔDEF <A ≅ <D <B ≅ <E <C ≅ <F AB ≅ DE BC ≅ EF 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. Since <CDB = <ABD, are corresponding parts of these 2 Δ’s, they must be equal. Therefore, <CDB = <ABD = 25° 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. Since <CDB = <ABD, are corresponding parts of these 2 Δ’s, they must be equal. Therefore, <CDB = <ABD = 25° 9 9 25° 17 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 So ΔABC ≅ ΔA’B’C’ by SAS. 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 Δ ACM ≅ Δ ABM. A 30° 5 5 B C

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 30° 5 5 M B C

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. AB ≅ AC BM ≅ CM AM ≅ AM so Δ ACM ≅ Δ ABM A 30° 5 5 M B C

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. AB ≅ AC BM ≅ CM AM ≅ AM so Δ ACM ≅ Δ ABM A 30° 5 5 M B 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 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 30° 5 5 M B C 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. 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 D C 23° B

Problem 3.17 AC = CD = DB, and < B = 23°. Find < A. < ADC = < DCB + < B = 46°. Since AC = DC, we have < A = < ADC = 46°. A 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 D C 23° B

Problem 3.18 XY = XZ = 8 and < X = 60°. Find YZ. X 60° 8 8 Y Z

Problem 3.18 Since <X = 60°, <Y + <Z = 180° - 60° = 120° Since XY = XZ, then <Y = <Z, so <Y = <Z = 60°. X 60° 8 8 Y Z

Problem 3.18 Therefore, all angles of Δ XYZ are equal. Specifically, <X = <Y means YZ = XZ, so YZ = 8. X 60° 8 8 Y Z

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. ΔPAB is an equilateral triangle and ABCD is a square. P 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 ABCD is a square. Draw BD so we can see <PBD. P A B D C

Problem 3.19 ΔPAB is an equilateral triangle and ABCD is a square. Since <BAD = 90°, the other 2 angles ΔBAD = (180 – 90)/2 = 45°. ΔPAB is an equilateral triangle and ABCD is a square. Draw BD so we can see <PBD. Since ΔPAB is equilateral, <PBA = 60°. ΔBAD is isosceles, since AB = AD. P A B D C

Problem 3.19 Therefore: <PBD = <PBA + <ABD = 105° Since <BAD = 90°, the other 2 angles ΔBAD = (180 – 90)/2 = 45°. ΔPAB is an equilateral triangle and ABCD is a square. Draw BD so we can see <PBD. Since ΔPAB is equilateral, <PBA = 60°. ΔBAD is isosceles, since AB = AD. P A B D C

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. 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. 24°

Problem 3.20 O is the center of the circle, AB || CD, <ABO = 24°, and <OBC = <OCD. Find <BOC. 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. 24° 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. 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. 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 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: The 2 angles given are equal. In this case, we have 3x + 4 = x + 17, so x = 6½°. 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: The 2 angles given are equal. In this case, we have 3x + 4 = x + 17, so x = 6½°. 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°. 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 3.44 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 <ACE = 90 - <BCE = <B/2.

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

Fini! The end of Chapter 3!