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TitleA Review of Geometry
TagsTriangle Circle Perpendicular Trigonometric Functions
File Size677.6 KB
Total Pages19
Document Text Contents
Page 1

A Review of Geometry

Khor Shi-Jie

May 17, 2012

Page 2

Contents

1 What’s Next? 2

2 Chasing Pavements 5
2.1 Your Toolbox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.1.1 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.1.2 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.1.3 Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 More Powerful Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3 Auxiliary Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1

Page 9

CHAPTER 2. CHASING PAVEMENTS 8

Corollary 2 In a triangle 4ABC and D is a point on BC. We have AB2 − BD2 =
AC2 − CD2 if and only if AD ⊥ BC.

Corollary 3 Given a quadrilateral ABCD, the diagonals AC and BD are perpendicular if
and only if AB2 + CD2 = AD2 + BC2.

Isosceles triangles and equilateral triangles often come out in problems too and one should
be familiar with its properties. For example, in an isosceles triangle, the altitude from the
vertex between the two equal side both a median and an angle bisector of the triangle. One
should also be aware of the likelihood of congruent triangles erected from the sides of an
equilateral triangle.

A cevian of a triangle is a segment connecting a vertex of the triangle to a point in the
opposite side of the triangle. Examples of cevians include the altitude of the triangle, the
median of the triangle and the angle bisector of the triangle. There is a very powerful theorem
which allows us to calculate the length of these cevians given the lengths of the sides of the
triangle.

Figure 2.1.2

Theorem 2 (Stewart’s Theorem) In triangle 4ABC, D is a point on BC. Suppose

AB = c, AC = b, BC = a,AD = d,BD = m,CD = n, we have d2 =
b2m + c2n

a
−mn.

Corollary 4 (Pappus’ Theorem) If D is the midpoint of BC, then we have d2 =
2b2 + 2c2 − a2

4
.

2.1.2 Circles

There are several essential circular properties that students must know (it is covered in Sec
2 syllabus). Here’s a list of these essential properties:

Property 1 Angles subtended by the same chord are equal i.e. ∠AP1B = ∠AP2B =
∠AP3B. Angles subtended by chords of equal length are equal.

Page 10

CHAPTER 2. CHASING PAVEMENTS 9

Figure 2.1.3

Property 2 Angle subtended by a diameter is 90◦.

Property 3 Angles at the centre of the circle is twice the angle at the opposite arc i.e.
∠AOB = 2∠ACB, reflex ∠AOB = 2∠ADB.

Figure 2.1.4

Property 4 Angles in opposite segments are supplementary i.e. ∠ABC + ∠ADC = 360◦.

Figure 2.1.5

Property 5 A tangent to the circle is perpendicular to the radius of the circle.

Property 6 The two tangents drawn from an external point to a circle is equal in length.

Page 18

CHAPTER 2. CHASING PAVEMENTS 17

20. (SMO(S) 2009 P16) 4ABC is a triangle and D is a point on side BC. Point E is on
side AB such that DE is the angle bisector of ∠ADB, and point F is on side AC such

that DF is the angle bisector of ∠ADC. Find the value of
AE

EB
·
BD

DC
·
CF

FA
.

21. (SMO(S) 2009 P29) ABCD is a rectangle, E is the midpoint of AD and F is the
midpoint of CE. If the area of triangle 4BDF is 12 cm2, find the area of rectangle
ABCD in cm2.

22. (SMO(O) 2009 P2) Let A1, A2, · · ·A6 be 6 points on a circle in this order such that
Â1A2 = Â2A3, Â3A4 = Â4A5, Â5A6 = Â6A1, where Â1A2 denotes the arc length of the
arc A1A2 etc. It is also known that ∠A1A3A5 = 72

◦. Find the size of ∠A4A6A2 in
degrees.

23. (SMO(O) 2009 P4) Let P1, P2, · · ·P41 be 41 distinct points on segment BC of a triangle
4ABC, where AB = AC = 7. Evaluate the sum

∑4
1i=1(AP

2
i + PiB · PiC).

24. (SMO(J) 2010 P14) In triangle 4ABC, AB = 32 cm, AC = 36 cm and BC = 44 cm.
If M is the midpoint of BC, find the length of AM in cm.

25. (SMO(J) 2010 P32) Given ABCD is a square. Points E and F lie on the side BC and
CD respectively, such that BE = CF = 1

3
AB. G is the intersection of BF and DE.

If
Area of ABGD

Area of ABCD
=

m

n
is in its lowest term, find the value of m + n.

26. (SMO(J) 2010 P1) Let the diagonals of the square ABCD intersect at S and let P be
the midpoint of AB. Let M be the intersection of AC and PD and N the intersection
of BD and PC. A circle is inscribed in the quadrilateral PMSN . Prove that the radius
of the circle is MP −MS.

27. (SMO(S) 2010 P10) Let ABCD be a trapezium with AD parallel to BC and ∠ADC =
90◦. Given that M is the midpoint of AB with CM = 13

2
cm and BC +CD+DA = 17

cm, find the area of trapezium ABCD in cm2.

28. (SMO(S) 2010 P22) Given a circle with diameter AB, C and D are points on the
circle on the same side of AB such thath BD bisects ∠CBA. The chords AC and BD
intersect at E. It is given that AE = 169 cm and EC = 119 cm. If ED = x cm, find
the value of x.

Page 19

CHAPTER 2. CHASING PAVEMENTS 18

29. (SMO(S) 2010 P1) In the triangle 4ABC with AC > AB, D is the foot of perpendicu-
lar from A onto BC and E is the foot of perpendicular from D onto AC. Let F be the
point on the line DE such that EF ·DC = BD ·DE. Prove that AF is perpendicular
to BF .

30. (SMO(J) 2011 P33) In the following diagram, ABCD is a square, BD ‖ CE and
BE = BD. Let ∠E = x◦. Find x.

31. (SMO(J) 2011 P2) Two circles Γ1,Γ2 with radii r1, r2 respectively, touch internally at
point P . A tangent parallel to the diameter through P touches Γ1 at R and intersects
Γ2 at M and N . Prove that PR bisects ∠MPN .

32. (SMO(S) 2011 P21) ABCD is a convex quadrilateral such that AC intersects BD at
the midpoint E of BD. The point H is the foot of perpendicular from A onto DE,
and H lies in the interior of the segment DE. Suppose ∠BCA = 90◦, CE = 12 cm,
EH = 15 cm, AH = 40 cm and CD = x cm. Find the value of x.

33. (SMO(O) 2011 P4) Given an isosceles triangle 4ABC with AB = AC and ∠A = 20◦.
The point D lies on AC such that AD = BC. Determine ∠ABD in degrees.

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