1. Using Analytic Approach to Prove Results Relating to Rectilinear Figures

2. Analytic Approach We call this the analytic approach.
Do you remember the deductive approach for proofs you learnt in S2? Here is an example.
Actually, we can also perform the proofs by introducing a rectangular coordinate system to the figures. y x
In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP. A O B P
Solution
AO = BO given
∠AOP = ∠BOP = 90 given
PO = PO common side
∴ △AOP  △BOP SAS
∴ AP = BP corr. sides,  △s

3. Introduce a rectangular coordinate system to the figure.
In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP. y x A O B P
Solution
Introduce a rectangular coordinate system to the figure.
With the coordinate system in red, we can set the coordinates of the vertices of △ABP easily.
Which coordinate system would you introduce? The orange one or the red one?
Note: In setting the coordinates of the vertices, they must satisfy all the conditions given in the question. A O B P A O B P

4. Introduce a rectangular coordinate system to the figure.
In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP. y x A O B P
(0, b)
Solution
Introduce a rectangular coordinate system to the figure.
(a, 0)
(a, 0)
Let a be the length of AO and b be the length of PO, then the coordinates of A, B and P are
With the coordinate system in red, we can set the coordinates of the vertices of △ABP easily.
Which coordinate system would you introduce? The orange one or the red one?
Note: In setting the coordinates of the vertices, they must satisfy all the conditions given in the question.
(a, 0),
(a, 0) and
(0, b) respectively.

5. In the figure, AOB is a straight line, AO = BO and PO ⊥ AB
In the figure, AOB is a straight line, AO = BO and PO ⊥ AB. Prove that AP = BP. y x A O B P
(0, b)
Solution 2 0) ( )] [0 b a AP - + =
(a, 0)
(a, 0) 2 b a + = 2 0) ( ) (0 b a BP - + = 2 ) ( b a + - = 2 b a + =
∴ AB = BP

6. Follow-up question
In the figure, AB and CD are perpendicular bisectors to each other. They intersect at E. Prove that ADBC is a rhombus. A D B C E
Solution
Let AE = BE = a and CE = DE = b. E x y
Introduce a rectangular coordinate system as shown in the figure.
C(0, b)
A(a, 0)
B(a, 0)
The coordinates of A, D, B and C are
A(a, 0),
B(a, 0),
C(0, b) and
D(0, b)
D(0, b)
respectively.

7. Follow-up question (cont’d)
In the figure, AB and CD are perpendicular bisectors to each other. They intersect at E. Prove that ADBC is a rhombus. A D B C E
Solution 2 b a AD - + = )] ( [ )
A(a, 0)
D(0, b)
B(a, 0)
C(0, b) E x y 2 b a + - = ) ( 2 b a + = 2 b a BD - + = )] ( [ ) 2 b a + =

8. Follow-up question (cont’d)
In the figure, AB and CD are perpendicular bisectors to each other. They intersect at E. Prove that ADBC is a rhombus. A D B C E
Solution 2 b a AC - + = ) ( )] [
A(a, 0)
D(0, b)
B(a, 0)
C(0, b) E x y 2 b a + = 2 b a BC - + = ) ( 2 b a + - = ) ( 2 b a + =

9. Follow-up question (cont’d)
In the figure, AB and CD are perpendicular bisectors to each other. They intersect at E. Prove that ADBC is a rhombus. A D B C E
Solution
∵ AD = BD = BC = AC
A(a, 0)
D(0, b)
B(a, 0)
C(0, b) E x y
∴ ADBC is a rhombus.