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The Mathematics Department
Stage : 2nd Prep
Date : / /
1st Term
Practice Sheet
The Isosceles Triangle
[1] Complete:
1] ∆ ABC, AB = AC, mA = 120º, so mC = …30…º
2] ∆ XYZ is isosceles, if mX = 100º, so mY = …80….º
3] ∆ ABC is isosceles right-angled at B, the measure of its base angle equals …45º
4] The measure of the base angle of an isosceles triangle is 50º, so the measure of
its vertex angle = …80..º
5] If the three angles in any triangle are congruent, then the triangle is
equilateral
6] In ∆XYZ, mX = 48º and mY = 84º, then the type of the triangle according to
its lengths is …isosceles…
7] DEF is a triangle in which DE = DF, then mE = m…F….
8] In the isosceles triangle, if the measure of the vertex angle is 40°, then the
measure of one of its base angle equals …70°…..
9] If two angles in a triangle are congruent, then the two opposite sides to these
angles are …equal in length…. and the triangle is …isosceles…..
10] If the measure of one angle in a right-angled triangle is 45°, then the triangle
is … isosceles …
11] If the measure of an angle in an isosceles triangle is 60°, then the triangle is
…equilateral triangle….
12] In ∆ABC, AB = AC and m A = 60°, if its perimeter equals 15 cm, then BC =
…5… cm
13] Using the given data for each figure, complete to fill the spaces under each
figure:
1]
3]
2]
\\
//
x
\\
x = 20°
x = 56°
5]
x
//
2x
62°
x = 50°
4]
\\
//
\\
50°
5x
x
6]
42°
//
y
x = 54° & y =
117°
\\
63°
//
y
\\
x
x = 69° & y = 111°
1
y
//
||
x
x = 60° & y = 120°
14] The number of symmetry axes of the equilateral triangle is …3..
15] The number of symmetry axes of the isosceles triangle is …1..
16] The number of symmetry axes of the scalene triangle is …zero..
17] The median of the isosceles triangle drawn from the vertex angle bisects
……bisects this angle…. and is …perpendicular to its base….
18] The symmetry axis of the line segment is the straight line that …is
perpendicular to it from its midpoint….
19] Any point belongs to the axis of a line segment is …equidistant.. from its
terminals
20] If the point C belongs to the symmetry axis of the line segment AB , then
…CA.. = …CB….
[2] Choose the correct answer:
1] The isosceles triangle has …. axis of symmetry.
i) one
ii) two
iii) three
iv) four
2] AB = AC and m (B) = 40º in the triangle ABC, then the exterior angle at its
vertex is ….. angle.
i) acute
ii) obtuse
iii) right
iv) straight
3] The measure of the exterior angle of the equilateral triangle = ……..º
i) 30
ii) 150
iii) 90
iv) 120
4] If ∆ ABC has only one axis of symmetry and mABC = 120º, then mA = …º
i) 60
ii) 120
iii) 30
iv) 40
5] The symmetric axis of the line segment is any line ……….
i) Perpendicular to it.
ii) bisects and perpendicular to it.
iii) cut the line segment.
iv) perpendicular to it at one of its two end points
[3] In the given figure:
AC = CB = CD
, mB = 36 , find mD
In ∆ ACB:
A
AC = CB Þ mCAB = mCBA = 36°
(exterior angle)
\
, mACD = 36° + 36° = 72°
36°
In ∆ ACD: CA = CD Þ
B
mCAD = mCDA = (180° -72°)  2 = 54°
2
\
C
\
D
[4] In the given figure:
AB = AC = CD, m BAC = 30
, m CAD = 64.
Find: mBCD
A
30° 64°
D
\\
//
\\
In ∆ ABC: AB = AC, m BAC = 30
mACB = (180 - 30)  2 = 75°
B
(1)
C
In ∆ CAD: CA = CD, m  CAD = 64
mCAD = mCDA = 64°
mACD = 180° - (64° + 64°) = 52°
mBCD = mBAC + mACD = 75°+52°=127°
[5] In the given figure:
AB = AC, DB = DC
m A = 80
mDBC = 30
Find: m ABC , mABD , m  BDC
A
80°
\\
//
D
/
\
In ∆ AB = AC: AB = AC Þ
B
mABC = mACB = (180° - 80°)  2 = 50°
30°
C
(1)
mABD = 50° - 30° = 20°
In ∆ DBC: DB = DC
mDBC = mDCB = 30
mBDC = 180° - (30° + 30°) = 120°
[6] In the given figure:
AB = AC, DE // BA
, m A = 34 Find: m ACB
and m CDE
E
C
In ∆ ABC:
\\
AB = AC and mA = 34
A
mACB = mB = (180° - 34°)  2 = 73°
DE // BA Þ
mD + mB = 180°
D
<
(interior angles)
mCDE = 180° - 73° = 107°
3
(1)
34°
\\ <
B
[7] In the given figure:
E  AB , AD = BC, mA = mB
& mDEC = 40°.
Find with proof: mEDC
C
D
=
=
In ∆ ADE , ∆ BCE:
ìï 1) AD = BC
ïï
ïí 2) AE = BE
ïï
ïïî 3) mÐA = mÐB
∆ ADE ≡ ∆ BCE
DE = CE Þ mEDC = mECD
In ∆ EDC:
mDEC = 40°
mEDC = (180 – 40)  2 = 70°
40°
B
●
|
[8] In the given figure:
AB = AC, and DB = DE
Prove that:
AC // ED
E
● A
|
A
E
C
/
B
/
//
\\
In ∆ ABC: AB = AC
mABC = mC
(1)
In ∆ DBC:
DB = DE Þ mDBE = mE
(2)
mABC = mDBE
(3)
From (1) , (2) and (3)
mC = mE
They are alternate angles Þ AC // ED
D
[9] In the given figure:
mABD = 110 and
mBAE = 140
Prove that:  ABC is an isosceles triangle
m (ABC) + m (ABD) = 180°
m (ABC) = 180° - 110 = 70°
m (BAC) + m (BAE) = 180
m (BAC) = 180° - 140° = 40°
In ∆ ABC: m (C) = 180° - (70 + 40) = 70°
m (C) = m (ABC) = 70°
AB = AC Þ  ABC is an isosceles triangle
4
C
B
110°
A
140°
D
E
[10] In the given figure:
DB = DC and
mABD = mACD
Prove that:  ABC is an isosceles.
A
B
In ∆ DBC: DB = DC
m (DBC) = m (DCB)
(1)
m ( ABD) = m (ACD)
(2)
(2) – (1) Þ m (ABC) = m (ACB)
AB = AC Þ
 ABC is an isosceles.
C
//
\\
D
[11] In the given figure:
AB = AC and
mABD = mACD
Prove that: DB = DC
A
\\
In ∆ABC: AB = AC
m(ABC) = m(ACB)
(1)
m ( ABD) = m (  ACD)
(2)
(1) – (2) Þ m(DBC) = m(DCB)
DB = DC
//
D
●
●
C
B
[12] In the given figure:
XY // AC , mABX = 62
and mC = 56
Prove that: AC = BC
A
X
62°
XY // AC
m(A) = m (ABX) = 62 (alternate angles)
In ∆ ABC:
m (ABC) = 180° - (62° + 56) = 62°
m(A) = m(ABC)
AC = BC
B
56°
C
Y
[13] In the given figure:
ABCD is a quadrilateral
AB = AD, AD // BC
,mABD = 36, mC = 72
Prove that: BC = BD
D
< \\
A
//
36°
In ∆ ABD:
AB = AD
m (ABD) = m(ADB) = 36
AD // BC
m(ADB) = m(DBC) = 36° (alternate angles)
In ∆ DBC:
m(BDC) = 180° - (72° + 36°) = 72°
m(BDC) = m(C)
BC = BD
5
C
72°
<
B
[14] In the given figure:
ABC is a right-angled triangle at B in which
mA = 45°, BD bisects B and AD = 20 cm.
i) Find the length of AC
ii) Prove that the triangle DBC is an
isosceles triangle
A
D
In  ABD
m (A) = 45°, m (B) = 90° and BD bisects B
m (ABD) = 90° ÷ 2 = 45°
m (ADB) = 180° - (45° + 45°) = 90°
BD  AC
In  ABC
m (C) = 180° - (45° + 90°) = 45°
 ABC is isosceles
BD  AC
→
D is a midpoint
AD = 20 cm
→
AC = 40 cm
C
[15] In the given figure:
ABCD is a trapezium in which AD // BC , and
AE bisects A. BD bisects B
Prove that:
i) AB = AD
ii) AE ^ BD
iii) BE = ED
D
In  ABD
m (ADB) = m (ABD)
AE bisects A
 ABD is isosceles
AE  DB
E is midpoint
C
(Alternate angles)
AB = AD
→
BE = ED
6
●
B
A
<
★
★
E
AD // BC
m (ADB) = m (DBC)
m (ABD) = m (DBC)
m (ADB) = m (ABD)
●
<
●
●
B