In the figure, Y is the orthocentre of triangle ABC. AY and CY are produced to meet BC and AB at M and N respectively. It is given that AB = AC = 13 cm and BC = 10 cm.

(a) (i) Prove that triangle ABN ~ triangle CBN

To prove that triangles \( \triangle ABN \) and \( \triangle CBN \) are similar, we need to demonstrate that they have the same angles.

### Given Information

- Triangle \( \triangle ABC \) is isosceles with \( AB = AC = 13 \, \text{cm} \) and \( BC = 10 \, \text{cm} \).
- \( Y \) is the orthocenter of \( \triangle ABC \).
- \( AY \) and \( CY \) are extended to meet \( BC \) and \( AB \) at \( M \) and \( N \) respectively.

### Step 1: Identify Key Angles

Since \( \triangle ABC \) is isosceles (\( AB = AC \)), the base angles are equal:

\[
\angle ABC = \angle ACB
\]

### Step 2: Use Orthocenter Properties

The orthocenter \( Y \) of a triangle is the point where the altitudes intersect. In \( \triangle ABC \):

- \( AY \) is perpendicular to \( BC \), so \( \angle AYB = 90^\circ \).
- \( CY \) is perpendicular to \( AB \), so \( \angle CYA = 90^\circ \).

### Step 3: Prove \( \triangle ABN \sim \triangle CBN \)

To prove similarity, we need to show that corresponding angles in \( \triangle ABN \) and \( \triangle CBN \) are equal.

#### 1. \( \angle ABN = \angle CBN \)

Since \( Y \) is the orthocenter and \( CY \) is the altitude, \( \angle CYA = 90^\circ \). Extending \( CY \) to \( N \), we have:

\[
\angle ABN = 90^\circ
\]

Similarly, since \( AY \) is the altitude from \( A \) and \( AY \) extended meets \( BC \) at \( M \):

\[
\angle CBN = 90^\circ
\]

Thus, \(\angle ABN = \angle CBN = 90^\circ\).

#### 2. \( \angle BAN = \angle BCN \)

Since \(\triangle ABC\) is isosceles with \(AB = AC\), we have:

\[
\angle BAC = \angle ACB
\]

Therefore, \(\angle BAN = \angle BCN\).

#### 3. \( \angle ANB = \angle CNB \)

As \( \angle ANB \) and \( \angle CNB \) are vertically opposite angles, they are equal:

\[
\angle ANB = \angle CNB
\]

### Conclusion

Since:

- \(\angle ABN = \angle CBN = 90^\circ\)
- \(\angle BAN = \angle BCN\)
- \(\angle ANB = \angle CNB\)

By the **Angle-Angle (AA) Criterion** for similarity, we have:

\[
\triangle ABN \sim \triangle CBN
\]

Thus, we have proven that \(\triangle ABN\) is similar to \(\triangle CBN\).

Would you like more details or have any questions?

Here are 8 related questions you might find interesting:

1. How is the orthocenter of a triangle constructed geometrically?
2. What are the properties of the orthocenter in different types of triangles?
3. Can you prove the similarity of triangles using the Side-Side-Side (SSS) criterion?
4. How do we prove that the altitudes of a triangle are concurrent at the orthocenter?
5. What is the significance of similar triangles in solving geometric problems?
6. How can we find the orthocenter of a triangle with given coordinates?
7. How do you determine if two triangles are congruent?
8. How can the orthocenter be used in real-world applications or problem-solving scenarios?

**Tip:** To quickly identify similar triangles, look for equal angles or proportional sides.

Learn how to prove the similarity of triangles ABN and CBN using the properties of an isosceles triangle, orthocenter, and the Angle-Angle (AA) Criterion for Similarity. Detailed step-by-step solution provided with clear geometric insights. Suitable for Grades 9-12 students studying geometry.

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