a lesson on geometrical proof

8/13/2019 A Lesson on Geometrical Proof

http://slidepdf.com/reader/full/a-lesson-on-geometrical-proof 1/4

of 4 Learning & Teaching KLAng Dec 2012

« From a geometrical proof to others… »

Background information:

The geometrical properties of circle have long been part of secondary Mathematics. In

this article, I intend to show how a lesson can be enriched with multiple proofs of the “right

angle in semi-circle” and an extension of lesson in proving of a point inside or outside a

circle.

A property of a circle – “right angle in semi-circle”:

AOB is a diameter of the circle. A, C , B are points on the circumference of the circle.

Then, 90 ACB

Proof #1:

Consider this as a special case of the property of “angle at the circumference is half that angle

at the centre of the circle”.

ACB and AOB are subtending the

same arc. ACB is an angle at the

circumference, AOB is an angle

at the centre of the circle.

180 AOB

By angle at the circumference is half that angle at the centre of the circle,

we have 902

1 AOB ACB .

This is the most commonly found proof in a math textbook.

O A B

C

O A B

C




Proof #2:

Extend a line from C through O to the circumference, P .

Let ACP and PCB .

By the property of “angle at the circumference is half

the angle at the centre of the circle”,

2 AOP and 2 POB .

Since AOB is a straight angle, 18022 .

90 . Hence 90 ACB

Proof #3:

Join a line from C to O. Let ACP and PCB .

By the property of exterior angle of a triangle,

2 AOC and 2COB .

Since AOB is a straight angle, 18022 .

90 . Hence 90 ACB

In this proof, we move away from the property of “angle at the circumference is half

the angle at the centre of the circle”. This is a triangle properties proof.

Proof #4:

Insert a congruent ABC as ' BAC as shown.

Join a line from C to C ’ . Let ACP and PCB .

ACBC ’ is a parallelogram. The diagonals AB and CC ’ divide

equally at O. Therefore, ACBC ’ is a rectangle.

Hence 90 ACB

O A B

C

θ

2θ

α

2α

P

O A B

C

θ

2θ

α

2α θ α

O A B

C

θ α

C ’

θ α




As demonstrated, even a simple circle property can be proven in multiple ways, hence a

source of enrichment of mathematical thinking and creativity.

Let study another property of circle that can be suitably adapted as an extension to a lesson.

To prove:

Given that 3 points, A, B, C on the circumference of a circle,

Point P is in the circle and on the same side of C by the line AB.

Prove that APB is larger than ACB .

Extends a line from A through P to Q on the circumference of the circle.

Join a line from Q to B.

ACB AQB , angles in the same segment.

PBQ AQB APB

PBQ ACB APB

Therefore, ACB APB .

To prove:

Given that 3 points, A, B, C on the circumference of a circle,

Point P is outside the circle and on the same side

of C by the line AB, as shown in the figure.

Prove that APB is smaller than ACB .

O

A

B

C

P•

C Q

O

A

B

P•

•

O

A

B

C P•




Extends a line from A through Q to P on the circumference of the circle.

Join a line from Q to B.

ACB AQB , angles in the same segment

QBP QPB AQB

QBP APB ACB

APBQBP ACB

Therefore, APB ACB

In conclusion:

The first activity provides students an opportunity to be creative and enriches the thinking

experience.

The second activity provides a challenge for students to explore the proving process.

C Q

O

A

B

P•

•

a lesson on geometrical proof

Documents