EXERCISE 7.2 In an isosceles triangle ABC , with $$ \mathrm{AB}=\mathrm{AC} $$, the bisectors of $$ \angle \mathrm{B} $$ and $$ \angle \mathrm{C} $$ intersect each other at O . Join A to O . Show that : $$ \begin{array}{ll} ext { (i) } \mathrm{OB}=\mathrm{OC} & ext { (ii) } \mathrm{AO} ext { bisects } \angle \mathrm{A}\end{array} $$

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EXERCISE 7.2 In an isosceles triangle ABC , with $$ \mathrm{AB}=\mathrm{AC} $$, the bisectors of $$ \angle \mathrm{B} $$ and $$ \angle \mathrm{C} $$ intersect each other at O . Join A to O . Show that : $$ \begin{array}{ll}	ext { (i) } \mathrm{OB}=\mathrm{OC} & 	ext { (ii) } \mathrm{AO} 	ext { bisects } \angle \mathrm{A}\end{array} $$

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To solve the problem, we will analyze the isosceles triangle $$ ABC $$ where $$ AB = AC $$ and the angle bisectors of $$ \angle B $$ and $$ \angle C $$ intersect at point $$ O $$. We need to show two things:

1. $$ OB = OC $$
2. $$ AO $$ bisects $$ \angle A $$

### Step 1: Understanding the Geometry

1. **Isosceles Triangle**: Since $$ AB = AC $$, we know that $$ \angle B = \angle C $$.
2. **Angle Bisectors**: The angle bisector of $$ \angle B $$ divides it into two equal angles, and similarly for $$ \angle C $$.

### Step 2: Proving $$ OB = OC $$

Since $$ O $$ is the intersection of the angle bisectors of $$ \angle B $$ and $$ \angle C $$, we can denote the angles as follows:

- Let $$ \angle ABC = \angle B = x $$
- Let $$ \angle ACB = \angle C = x $$

Thus, the angle bisectors create two angles at point $$ O $$:

- $$ \angle OBC = \frac{x}{2} $$
- $$ \angle OCB = \frac{x}{2} $$

Now, consider triangle $$ OBC $$:

- The angles in triangle $$ OBC $$ are:
  - $$ \angle OBC = \frac{x}{2} $$
  - $$ \angle OCB = \frac{x}{2} $$
  - $$ \angle BOC = 180^\circ - \left(\frac{x}{2} + \frac{x}{2}\right) = 180^\circ - x $$

Since $$ \angle OBC = \angle OCB $$, triangle $$ OBC $$ is isosceles, which implies:

$$
OB = OC
$$

### Step 3: Proving $$ AO $$ Bisects $$ \angle A $$

Next, we need to show that line $$ AO $$ bisects $$ \angle A $$.

1. **Angles at A**: The angle $$ \angle A $$ can be expressed as:
   $$
   \angle A = 180^\circ - (x + x) = 180^\circ - 2x
   $$

2. **Using the Angle Bisector Theorem**: Since $$ O $$ is the intersection of the angle bisectors, we can analyze the angles formed at point $$ A $$:
   - The angle $$ \angle OAB $$ and $$ \angle OAC $$ must be equal because $$ AO $$ is the angle bisector of $$ \angle A $$.

3. **Calculating the Angles**:
   - Since $$ \angle OAB + \angle OAC = \angle A $$ and they are equal, we can denote:
   $$
   \angle OAB = \angle OAC = \frac{1}{2} \angle A = \frac{1}{2}(180^\circ - 2x) = 90^\circ - x
   $$

Thus, we conclude that $$ AO $$ bisects $$ \angle A $$.

### Conclusion

We have shown that:

1. $$ OB = OC $$
2. $$ AO $$ bisects $$ \angle A $$

This completes the proof for the given problem.
In triangle ABC, where AB = AC, the angle bisectors of angles B and C intersect at point O. 1. $$ OB = OC $$ 2. $$ AO $$ bisects angle A

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