Question

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that 

(i) AD bisects BC 

(ii) AD bisects ∠A.

Answer 1

Given: 

- Triangle \( \triangle ABC \) is an isosceles triangle where \( AB = AC \).
- \( AD \) is the altitude of \( \triangle ABC \) drawn from vertex \( A \) to the base \( BC \).

To prove:

(i) \( AD \) bisects \( BC \)
(ii) \( AD \) bisects \( \angle A \)

Proof for (i): \( AD \) bisects \( BC \)

1. Since \( AB = AC \), triangle \( \triangle ABC \) is isosceles, and by the definition of an isosceles triangle, the altitude from the vertex (in this case, \( A \)) will also act as the perpendicular bisector of the base \( BC \).
2. Therefore, the altitude \( AD \) divides the base \( BC \) into two equal parts:

\[ BD = DC \]

1. Since \( AD \) is both the altitude and the bisector of \( BC \), we can conclude that:

\[ \boxed{AD \text{ bisects } BC.} \]

Proof for (ii): \( AD \) bisects \( \angle A \)

1. In an isosceles triangle, the altitude drawn from the vertex (here, \( A \)) to the base (\( BC \)) not only bisects the base but also bisects the vertex angle.
2. Since \( AB = AC \) and \( AD \) is perpendicular to \( BC \), triangles \( \triangle ABD \) and \( \triangle ACD \) are congruent by the hypotenuse-leg congruence criterion (i.e., \( AB = AC \), \( AD = AD \), and \( BD = DC \)).
3. Since these two triangles are congruent, the corresponding angles at \( A \) are equal, implying:

\[ \angle ABD = \angle ACD \]

1. Thus, the altitude \( AD \) bisects \( \angle A \).
2. Therefore, we conclude:

\[ \boxed{AD \text{ bisects } \angle A.} \]

Conclusion:

We have shown that in an isosceles triangle \( ABC \) where \( AB = AC \), the altitude \( AD \) has two properties:

1. \( AD \) bisects the base \( BC \).
2. \( AD \) bisects the vertex angle \( \angle A \).