ANH 7.

PRACTICE THE FIRST CONGRUENCE OF TRIANGLES SIDE-SIDE-SIDE(S.S.S)

NEW WORK
+ triangle:tamgiác
+ segment: đoạnthẳng
+ arc:cungtròn
+ half-plane:nửamặtphẳng
+ we obtain: ta được
+ congruent:bằngnhau
+ corresponding angle: góctươngứng
+ measure:đo
+ measurement:sốđo
+ length:độdài
+ prove: chứngminh
+ common side:cạnhchung
+ consider:xét
+ the bisector: tiaphângiác
+ radius: bánkính
+ center: tâm
II- Exercises
Exercise 19:(Textbook-page114)
a) Consider
ADE and
BDE have:
AD = BD (given)
DE is common side
AE = BE (given)


ADE =
BDE (S.S.S)
b) According a) we have:
ADE =
BDE


DAE =
DBE (two corresponding angles)
Exercise 28:(workbook-page140)
Consider
CAD and
CBD have:
AC = BC = 3cm (given)
CD is common side
AD = BD = 2cm(given)


CAD =
CBD (S.S.S)


CAD =
CBD (two corresponding angles)
Exercise 32:(workbook-page141)
Consider
AMB and
AMC have:
AB = AC (given)
AM is common side
MB = MC (Since M is the midpointof BC)


AMB =
AMC (S.S.S)


AMB =
AMC (two corresponding angles)
That AMB + AMC = 1800 (two linear angles)

AMB = AMC = 900

AM
BC at M
Exercise 3.2:(workbook-page142)
a) - Draw the segment BC = 2cm
- In the same side of the half-plane with edge BC, draw arc center B radius 3cm and arc center C radius 3cm. Two arcs above cut each other at A
- Draw the segments AB, AC, we obtain triangle ABC
b) Consider
AEB and
AEC have:
AB = AC = 3cm (given)
AE is common side
EB = EC (Since E is the midpoint of BC)


AEB =
AEC (S.S.S)


BAE =
CAE (two corresponding angles)

AE is the bisector of angle BAC
Do exercises 33,34,3.4 (Workbook-page 141,142)







PRACTICE THE SECOND CONGRUENCE OF TRIANGLES SIDE-ANGLE-SIDE(S.A.S)

NEW WORK
+ included angle: góc xen giữa
+ corollary: hệquả
+ two legs:haicạnhgócvuông
+ midpoint:trungđiểm
+Two vertical angles:haigócđốiđỉnh
+ two linear angles:haigóckềbù
+ two alternate angles:haigóc sole trong
+ exterior angle:gócngoài
+ interior angle:góctrong
+ two corresponding angles:haigócđồngvị
+ two same-side interior angles:haigóctrongcùngphía
+ two corresponding sides:haicạnhtươngứng
II- Exercises
Exercise 29:(Textbook-page 120)
We have: AC = AD + DC (D
AC)
AE = AB + BE (B
AE)
That AD = AB (given); DC = BE (given)

AC = AE
Consider
ABC and
ADE have:
AB = AD (given)

A is common angle

AC = AE (cmt)


ABC =
ADE (S.A.S)
Exercise 40:(Workbook-page 142)

Consider
AKM and
BKM have:

AMK = BMK (= 900)
KM is common side
AM= BM (Since E is the midpoint of BC)


AKM =
BKM (S.A.S)

AKM = BKM (Two corresponding angles)

KM is the bisector of angle AKB
Exercise 42:(Workbook-page 142)

Consider
ABC and
DEC have:

ACB = DCE (Two vertical angles )
AC = DC (given)
BC = EC (given)


ABC =
DEC (S.A.S)

A = D (Two corresponding angles)
That A = 900 hence D = 900
Exercise 43:(Workbook-page 142)

a) Consider
ABD and
EBD have:

AB = EB(given )
BD is common side
ABD = EBD (since BDthe bisector of angle ABC)


ABD =
EBD (S.A.S)

DA = DE (Two corresponding sides)
b) According a) we have:
ABD =
EBD

A = BED (Two corresponding angles)
That A = 900 hence BED = 900
Exercise 44:(Workbook-page 142)