Sec 4 · Additional Mathematics

Proofs

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1. Proving trigonometric identities

Step	How
1	start from the more complicated side
2	write $\tan$ , $\sec$ , $\operatorname{cosec}$ , $\cot$ in $\sin$ and $\cos$
3	make one fraction; use $\sin^{2} x + \cos^{2} x = 1$
4	end with the other side: $= \text{RHS}$

Prove (1 − cos²x)(1 + tan²x) = tan²x

\displaystyle \text{LHS} = (1 - \cos^{2} x)(1 + \tan^{2} x)

e.g.

\displaystyle (1 - \cos^{2} x)(1 + \tan^{2} x)

\displaystyle = \sin^{2} x \sec^{2} x

\displaystyle = \dfrac{\sin^{2} x}{\cos^{2} x}

\displaystyle = \tan^{2} x

ends at the right-hand side

Prove (sin 2x) ÷ (1 + cos 2x) = tan x

\displaystyle \text{LHS} = \dfrac{\sin 2x}{1 + \cos 2x}

e.g.

\displaystyle \dfrac{\sin 2x}{1 + \cos 2x}

\displaystyle = \dfrac{2\sin x\cos x}{1 + 2\cos^{2} x - 1}

\displaystyle = \dfrac{2\sin x\cos x}{2\cos^{2} x}

\displaystyle = \tan x

ends at the right-hand side

📖 Worked example

Prove that $\dfrac{2\sin x\cos x}{\cos^{2} x - \sin^{2} x} = \tan 2x$ .

Step 1 of 4

Start from the more complicated side, the left-hand side

\text{LHS} = \dfrac{2\sin x\cos x}{\cos^{2} x - \sin^{2} x}

Practise this →

2. Writing a geometric proof

Statement	Reason to write
$\angle AXB = \angle CXD$ , lines crossing at $X$	vert. opp. ∠s
$\angle BAC = \angle ACD$ , $AB \parallel CD$	alt. ∠s, $AB \parallel CD$
$\angle ADE = \angle ABC$ , $DE \parallel BC$	corr. ∠s, $DE \parallel BC$
the same angle in both triangles	common ∠
the same side in both triangles	common side
$\angle ABC = \angle ACB$ when $AB = AC$	base ∠s of isos. △
$OA = OB$ , $O$ the centre	radii
$\angle OAT = 90^\circ$ , $AT$ a tangent	tangent ⊥ radius
$PA = PB$ , tangents from $P$	tangents from ext. point
$\angle ACB = 90^\circ$ , $AB$ a diameter	rt. ∠ in semicircle
$\angle ACB = \angle ADB$ , on the same arc $AB$	∠s in same segment
$\angle BAT = \angle ACB$ , $AT$ a tangent	alt. segment theorem
$MN \parallel BC$ , $MN = \dfrac{1}{2}BC$	midpoint theorem
$\angle ACD = \angle ABC + \angle BAC$ , $BC$ produced to $D$	ext. ∠ of △
$\angle A + \angle B + \angle C = 180^\circ$	∠ sum of △
$\angle ABD + \angle DBC = 180^\circ$ , $ABC$ a straight line	adj. ∠s on a str. line
$\angle AOB = 2\angle ACB$ , $O$ the centre	∠ at centre = 2 ∠ at circumference
$\angle ABC + \angle ADC = 180^\circ$ , $ABCD$ cyclic	∠s in opp. segments
$\angle DCE = \angle BAD$ , $BC$ produced to $E$ , $ABCD$ cyclic	ext. ∠ of cyclic quad.
$AM = MB$ , $OM \perp AB$ , $O$ the centre	⊥ from centre bisects chord

	Test	Equal parts
congruent $\equiv$	SSS	three pairs of sides
	SAS	two pairs of sides and the angle between them
	ASA / AAS	two pairs of angles and a pair of corresponding sides
	RHS	right angle, hypotenuse and one side
similar	AA	two pairs of angles
	SAS	two pairs of sides in the same ratio, the angle between them equal
	SSS	three pairs of sides in the same ratio

📖 Worked example

$AB$ is parallel to $DC$ . The lines $AD$ and $BC$ cross at $M$ , and $M$ is the mid-point of $AD$ . Prove that $\triangle ABM \equiv \triangle DCM$ , and state the congruence test.

Step 1 of 2

Match three pairs of parts, each with its reason

\angle BAM = \angle CDM

(alt. ∠s (AB ∥ DC))

AM = DM

(given)

\angle AMB = \angle DMC

(vert. opp. ∠s)

📖 Worked example

In triangle $ABC$ , $D$ lies on $AB$ and $E$ lies on $AC$ such that $\angle ADE = \angle ACB$ . $AD = 3$ cm, $DB = 2$ cm and $AC = 5$ cm. Prove that $\triangle ADE$ is similar to $\triangle ACB$ . Hence find $AE$ .

Step 1 of 5

Two pairs of equal angles, each with its reason

\angle ADE = \angle ACB

(given)

\angle DAE = \angle CAB

(common ∠)

Practise this →

3. The midpoint theorem

Midpoint theorem

\displaystyle MN \parallel BC,\quad MN = \dfrac{1}{2}BC

Triangle ABC with M the midpoint of AB and N the midpoint of AC; MN is parallel to BC.

$M$ , $N$ the midpoints of $AB$ , $AC$
Reason: midpoint theorem

Midpoints of a quadrilateral

\displaystyle PQ = \dfrac{1}{2}AC,\quad SR = \dfrac{1}{2}AC

Quadrilateral ABCD with P, Q, R, S the midpoints of its sides; PQ and SR are both parallel to the diagonal AC.

$PQ \parallel SR$ and $PQ = SR$ : $PQRS$ is a parallelogram

Diagonals of $ABCD$	$PQRS$ is a
neither equal nor perpendicular	parallelogram
equal (rectangle, isosceles trapezium)	rhombus
perpendicular (rhombus, kite)	rectangle
equal and perpendicular (square)	square

📖 Worked example

In triangle $ABC$ , $M$ is the mid-point of $AB$ and $N$ is the mid-point of $AC$ . $\angle ABC = 62^\circ$ . Show that $MN$ is parallel to $BC$ and $MN = \dfrac{1}{2}BC$ . Hence find $\angle AMN$ .

Step 1 of 2

M and N are the midpoints of two sides of triangle ABC

AM = MB

and

AN = NC

(given)

MN \parallel BC

and

MN = \dfrac{1}{2}BC

(midpoint theorem)

📖 Worked example

$ABCD$ is a rectangle. $P$ , $Q$ , $R$ and $S$ are the mid-points of $AB$ , $BC$ , $CD$ and $DA$ . Prove that $PQRS$ is a parallelogram, and state the most specific name for the quadrilateral $PQRS$ .

Step 1 of 6

Join AC. In triangle ABC, P and Q are midpoints

PQ \parallel AC

and

PQ = \dfrac{1}{2}AC

(midpoint theorem)

Practise this →

4. The tangent-chord theorem

Tangent-chord (alternate segment) theorem

\displaystyle \angle BAT = \angle ACB

Circle with tangent SAT at A and chord AB; angle BAT and angle ACB, with C on the far arc, are both marked x.

$C$ in the segment on the OTHER side of $AB$
Reason: alt. segment theorem

Tangent and secant

\displaystyle PT^{2} = PA \times PB

Tangent PT and a line PAB through the circle; angle PTA and angle PBT are marked equal.

$\triangle PTA \sim \triangle PBT$ (AA)

Intersecting chords

\displaystyle AX \times XC = BX \times XD

Chords AC and BD meeting at X; angle BAX and angle CDX are marked equal.

$\triangle AXB \sim \triangle DXC$ (AA)

📖 Worked example

$P$ , $Q$ , $R$ and $S$ lie on the circle and $YPX$ is the tangent to the circle at $P$ . $\angle QPX = 62^\circ$ . Find $\angle PSQ$ , giving a reason for each step.

Step 1 of 2

Compare the angle between the tangent and the chord PQ with the angle at R

\angle PRQ = \angle QPX

(alt. segment theorem)

\angle PRQ = 62^\circ

📖 Worked example

$PT$ is the tangent to the circle at $T$ , and the straight line $PAB$ cuts the circle at $A$ and $B$ . $PA = 4$ cm and $AB = 5$ cm. Prove that $\triangle PTA$ is similar to $\triangle PBT$ , and hence that $PT^{2} = PA \times PB$ . Find $PT$ .

Step 1 of 5

Two pairs of equal angles, each with its reason

\angle TPA = \angle BPT

(common ∠)

\angle PTA = \angle PBT

(alt. segment theorem)

Practise this →

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Sec 4 · Additional Mathematics

Proofs

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Start free Worksheet (PDF)

1. Proving trigonometric identities

Step	How
1	start from the more complicated side
2	write $\tan$ , $\sec$ , $\operatorname{cosec}$ , $\cot$ in $\sin$ and $\cos$
3	make one fraction; use $\sin^{2} x + \cos^{2} x = 1$
4	end with the other side: $= \text{RHS}$

Prove (1 − cos²x)(1 + tan²x) = tan²x

\displaystyle \text{LHS} = (1 - \cos^{2} x)(1 + \tan^{2} x)

e.g.

\displaystyle (1 - \cos^{2} x)(1 + \tan^{2} x)

\displaystyle = \sin^{2} x \sec^{2} x

\displaystyle = \dfrac{\sin^{2} x}{\cos^{2} x}

\displaystyle = \tan^{2} x

ends at the right-hand side

Prove (sin 2x) ÷ (1 + cos 2x) = tan x

\displaystyle \text{LHS} = \dfrac{\sin 2x}{1 + \cos 2x}

e.g.

\displaystyle \dfrac{\sin 2x}{1 + \cos 2x}

\displaystyle = \dfrac{2\sin x\cos x}{1 + 2\cos^{2} x - 1}

\displaystyle = \dfrac{2\sin x\cos x}{2\cos^{2} x}

\displaystyle = \tan x

ends at the right-hand side

📖 Worked example

Prove that $\dfrac{2\sin x\cos x}{\cos^{2} x - \sin^{2} x} = \tan 2x$ .

Step 1 of 4

Start from the more complicated side, the left-hand side

\text{LHS} = \dfrac{2\sin x\cos x}{\cos^{2} x - \sin^{2} x}

Practise this →

2. Writing a geometric proof

Statement	Reason to write
$\angle AXB = \angle CXD$ , lines crossing at $X$	vert. opp. ∠s
$\angle BAC = \angle ACD$ , $AB \parallel CD$	alt. ∠s, $AB \parallel CD$
$\angle ADE = \angle ABC$ , $DE \parallel BC$	corr. ∠s, $DE \parallel BC$
the same angle in both triangles	common ∠
the same side in both triangles	common side
$\angle ABC = \angle ACB$ when $AB = AC$	base ∠s of isos. △
$OA = OB$ , $O$ the centre	radii
$\angle OAT = 90^\circ$ , $AT$ a tangent	tangent ⊥ radius
$PA = PB$ , tangents from $P$	tangents from ext. point
$\angle ACB = 90^\circ$ , $AB$ a diameter	rt. ∠ in semicircle
$\angle ACB = \angle ADB$ , on the same arc $AB$	∠s in same segment
$\angle BAT = \angle ACB$ , $AT$ a tangent	alt. segment theorem
$MN \parallel BC$ , $MN = \dfrac{1}{2}BC$	midpoint theorem
$\angle ACD = \angle ABC + \angle BAC$ , $BC$ produced to $D$	ext. ∠ of △
$\angle A + \angle B + \angle C = 180^\circ$	∠ sum of △
$\angle ABD + \angle DBC = 180^\circ$ , $ABC$ a straight line	adj. ∠s on a str. line
$\angle AOB = 2\angle ACB$ , $O$ the centre	∠ at centre = 2 ∠ at circumference
$\angle ABC + \angle ADC = 180^\circ$ , $ABCD$ cyclic	∠s in opp. segments
$\angle DCE = \angle BAD$ , $BC$ produced to $E$ , $ABCD$ cyclic	ext. ∠ of cyclic quad.
$AM = MB$ , $OM \perp AB$ , $O$ the centre	⊥ from centre bisects chord

	Test	Equal parts
congruent $\equiv$	SSS	three pairs of sides
	SAS	two pairs of sides and the angle between them
	ASA / AAS	two pairs of angles and a pair of corresponding sides
	RHS	right angle, hypotenuse and one side
similar	AA	two pairs of angles
	SAS	two pairs of sides in the same ratio, the angle between them equal
	SSS	three pairs of sides in the same ratio

📖 Worked example

$AB$ is parallel to $DC$ . The lines $AD$ and $BC$ cross at $M$ , and $M$ is the mid-point of $AD$ . Prove that $\triangle ABM \equiv \triangle DCM$ , and state the congruence test.

Step 1 of 2

Match three pairs of parts, each with its reason

\angle BAM = \angle CDM

(alt. ∠s (AB ∥ DC))

AM = DM

(given)

\angle AMB = \angle DMC

(vert. opp. ∠s)

📖 Worked example

Step 1 of 5

Two pairs of equal angles, each with its reason

\angle ADE = \angle ACB

(given)

\angle DAE = \angle CAB

(common ∠)

Practise this →

3. The midpoint theorem

Midpoint theorem

\displaystyle MN \parallel BC,\quad MN = \dfrac{1}{2}BC

$M$ , $N$ the midpoints of $AB$ , $AC$
Reason: midpoint theorem

Midpoints of a quadrilateral

\displaystyle PQ = \dfrac{1}{2}AC,\quad SR = \dfrac{1}{2}AC

$PQ \parallel SR$ and $PQ = SR$ : $PQRS$ is a parallelogram

Diagonals of $ABCD$	$PQRS$ is a
neither equal nor perpendicular	parallelogram
equal (rectangle, isosceles trapezium)	rhombus
perpendicular (rhombus, kite)	rectangle
equal and perpendicular (square)	square

📖 Worked example

In triangle $ABC$ , $M$ is the mid-point of $AB$ and $N$ is the mid-point of $AC$ . $\angle ABC = 62^\circ$ . Show that $MN$ is parallel to $BC$ and $MN = \dfrac{1}{2}BC$ . Hence find $\angle AMN$ .

Step 1 of 2

M and N are the midpoints of two sides of triangle ABC

AM = MB

and

AN = NC

(given)

MN \parallel BC

and

MN = \dfrac{1}{2}BC

(midpoint theorem)

📖 Worked example

Step 1 of 6

Join AC. In triangle ABC, P and Q are midpoints

PQ \parallel AC

and

PQ = \dfrac{1}{2}AC

(midpoint theorem)

Practise this →

4. The tangent-chord theorem

Tangent-chord (alternate segment) theorem

\displaystyle \angle BAT = \angle ACB

$C$ in the segment on the OTHER side of $AB$
Reason: alt. segment theorem

Tangent and secant

\displaystyle PT^{2} = PA \times PB

$\triangle PTA \sim \triangle PBT$ (AA)

Intersecting chords

\displaystyle AX \times XC = BX \times XD

$\triangle AXB \sim \triangle DXC$ (AA)

📖 Worked example

$P$ , $Q$ , $R$ and $S$ lie on the circle and $YPX$ is the tangent to the circle at $P$ . $\angle QPX = 62^\circ$ . Find $\angle PSQ$ , giving a reason for each step.

Step 1 of 2

Compare the angle between the tangent and the chord PQ with the angle at R

\angle PRQ = \angle QPX

(alt. segment theorem)

\angle PRQ = 62^\circ

📖 Worked example

Step 1 of 5

Two pairs of equal angles, each with its reason

\angle TPA = \angle BPT

(common ∠)

\angle PTA = \angle PBT

(alt. segment theorem)

Practise this →

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