1. Important Lines in a Triangle

2. Important Lines in a Triangle
In △ABC,
AD bisects BAC, i.e. BAD = DAC,
BE bisects ABC, i.e. ABE = EBC,
CF bisects ACB, i.e. ACF = FCB. A B C D F E
 Angle bisectors must lie inside a triangle.
We called AD, BE and CF the angle bisectors of △ABC.
A line segment that bisects an interior angle in a triangle is called an angle bisector of the triangle.

3. We called AD, BE and CF the medians of △ABC.
AD bisects BC, i.e. BD = DC,
BE bisects AC, i.e. AE = EC,
CF bisects AB, i.e. AF = FB.
In △ABC, A B C D
 Medians must lie inside a triangle. E F
We called AD, BE and CF the medians of △ABC.
A line segment joining a vertex of a triangle to the mid-point of its opposite side is called a median of the triangle.

4. We called AD, BE and CF the altitudes of △ABC.
AD ⊥ BC,
BE ⊥ AC,
CF ⊥ AB.
In △ABC, A B C
 Altitudes may lie inside or outside a triangle. In some cases, it may even coincide with the side of a triangle. F D E
We called AD, BE and CF the altitudes of △ABC.
A line segment drawn from a vertex of a triangle perpendicular to the opposite side of the vertex is called an altitude of the triangle.

5. We call MN, PQ and ST the perpendicular bisectors of △ABC.
In △ABC,
MN bisects BC and MN ⊥ BC.
PQ bisects AC and PQ ⊥ AC.
ST bisects AB and ST ⊥ AB. A B C S T P Q M N
We call MN, PQ and ST the perpendicular bisectors of △ABC.
A perpendicular line that bisects a side of a triangle is called a perpendicular bisector of the triangle.

6. What is the name of the orange line in each of the following triangles?
(b)
altitude
angle bisector
perpendicular bisector
(c)
(d)
median

7. Follow-up question A B C D
In the figure, DB and DC are the angle bisectors of ABC and ACB respectively, ABC = 54 and ACB = 60. Find BDC.
Solution
∵ DB is the angle bisector of ABC.
∴ DBC = ABD
i.e. DBC =  ABC
=  54
= 27

8. Follow-up question (cont’d)A B C D
In the figure, DB and DC are the angle bisectors of ABC and ACB respectively, ABC = 54 and ACB = 60. Find BDC.
Solution
∵ DC is the angle bisector of ACB.
∴ DCB = ACD
i.e. DCB =  ACB
=  60
= 30

9. Follow-up question (cont’d)A B C D
In the figure, DB and DC are the angle bisectors of ABC and ACB respectively, ABC = 54 and ACB = 60. Find BDC.
Solution
Consider △ABC.
DBC + DCB + BDC = 180
( sum of △)
27 + 30 + BDC = 180
BDC = 123