Page 38 - Mathematics_Form_Two

Congruence

Figure 2.7 shows that, Proof: From ADB and ADC, , it

AB = PQ (given) implies that

AC = PR (given) AB = AC (given)
ˆ
ˆ
ˆ
ˆ
ABC = PQR = 90° (given). ADB = ADC = 90° (given)
FOR ONLINE READING ONLY
AD AD= is a common side.
Mathematics for Secondary Schools fit exactly. Thus, Figure 2.7 shows that hence DB DC= (definition of
If two triangles satisfy the conditions in

the RHS postulate, then the triangles must
ADC (by RHS),
Therefore, ADB

PQR (by RHS).

ABC 
congruence of triangles).
Since the two triangles in Figure 2.7 are
congruent, then, all the corresponding
sides and angles are equal. That is,
In the following figure, point R is
BC = QR Example 2�11
equidistant from two lines l and
ˆ
ˆ
1
BAC = QPR l , which intersect at T. Prove that
2
ˆ
ˆ
ˆ
ˆ
BCA = QRP . RTV = RTS.
Example 2�10
Use the following figure to prove that
 ADB  ADC and DB = DC.

Solution:
ˆ
ˆ
Required to prove that RTV = RTS.
Proof: From RVT∆ and SRT∆ , it
follows that
RV = RS (given)
ˆ
ˆ
RVT = RST (given)
RT is a common side.
Solution Thus, ΔRVT ≅ ΔSRT (by RHS).
Given ABC, such that AD is
perpendicular to BC . Since ΔRVT ≅ ΔSRT , it follows that all
Required to prove that corresponding angles and sides are equal.

ˆ
(a) ADB  ADC Therefore, RTV=RTS.
ˆ
(b) DB BC=

32
Student's Book Form Two

11/10/2024 20:11:23
MATHEMATIC F2 v5.indd 32
MATHEMATIC F2 v5.indd 32 11/10/2024 20:11:23