Download Deductive Proof - Egyptian Language School

The Mathematics Department
Stage : 1st Prep
Date
: / /
2nd Term
Exercises on
Deductive Proof
Remarks:
Theorem (1):
If two straight lines intersect, then each two vertically
opposite angles are
equal in measure.
Theorem (2):
The sum of the measure of accumulative angles at a point
is equal to 360°.
Theorem (3):
The sum of the measure of interior angles of a triangle is
180°.
[1] Complete:
1] The sum of the measures of the interior angle of a triangle
= …180…..°
2] The sum of the measure of accumulative angles at a point is
……360°…….
[2] Choose the correct answer:
1] If two adjacent angles are complementary, then their outer
sides are…….
i) perpendicular
ii) intersection
iii) equal
iv) parallel
2] If the measures of two angles of a triangle are 55 and
45, then the triangle is ………..
i) right – angled
ii) acute – angled
iii) obtuse- angled
iv) equilateral
[3] In the opposite figure:
C
mAMB = 50°, mEMD = 80°, MC
bisects BMD and mCMD = 65°.
find: m (AME)
D
B
65°● ● 50°
80° M
Given: mAMB = 50°, mEMD = 80°,
MC bisects BMD , mCMD = 65°.
E
R.T.P.: m (AME)
Proof:
m(AMB)  50
3] If two straight lines intersecting, then every two opposite
angles are ……equal in measure…..
m(EMD)  80o
4] If two adjacent angles are supplementary, then their outer
sides are …on the same straight line……
MC bi sec ts BMD
5] If the two outer sides of two adjacent angles are
perpendicular then they are ……complementary……..
6] If two angles of a triangle are complementary, then the
triangle is ……a right-angled triangle….
m(CMD)  65o
 m(BMC)  65o
 m(AME)  360  (50  65  65  80)  100o
A
[4] In the opposite figure:
DE // BC , A ∈ DE , m (DAB) =
E
80° and m (EAC) = 50°.
Find the measures of the angles of
∆ ABC
Given: DE // BC , A ∈ DE , m
(DAB) = 80° , m (EAC) = 50°.
A
< 50° 80°
[6] In the opposite figure:
B ∈ AC , mCBE = 116°
D
BD bisects ABE
Find mABD
<
C
&
●
●
A
116°
B
C
Given: B ∈ AC , mCBE = 116°, BD bisects ABE
R.T.P.: mABD
B
Proof:
R.T.P.: angles of ∆ABC
Proof:
DE // BC
E
D
B  AC
m(CBE)  116o
A  DE
BD bi sec ts ABE
m(DAB)  80o & m(EAC)  50o
 m(B)  m(DAB) = 80o
 m(C)  m(EAC)  50o
 m(ABE)  180  116  64o
(alternate)
 m(ABD)  64  2  32o
(alternate)
 m(CAB)  180  (50  80)  50o
[7] In the opposite figure:
AC
[5] In the opposite figure:
mABC = 110°,mCBD =
35° and mABE = 140°
Find: mEBD
D
Given: AC
35°
Given: mABC =110°, mCBD = 35°,
mABE = 140°
R.T.P.: mEBD
110°
B 140°
A
o
m(ABC)  110 & m(CBD)  35 & m(ABE)  140
 m(EBD)  360  (110  35  140)  75
(Accumulative angles)
DE = {B},
mABD = 40°, BE bisects CBF
R.T.P.: mABF
B  AC
o
o
40°
A
Proof:
E
Proof:
o
DE = {B}, mABD = 40°
BE bisects CBF. Find mABF
C
D
m(ABD)  40
 m(CBE)  m(ABD)  400 (V.O.A)
BE bi sec ts CBF
 m(EBF)  m(CBE)  400
 m(ABF)  180  (40  40)  100o
B
●
●
F
C
E