If `a cos A =b cos B, " then " DeltaABC` is

Options

A. isosceles
B. right angled
C. equilateral
D. right angled isosceles

Answer

To solve the problem where \( a \cos A = b \cos B \) in triangle \( ABC \), we need to analyze the implications of this equation.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ a \cos A = b \cos B \]

2. **Use the cosine rule:**
The cosine rule states:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

3. **Substitute the cosine values into the equation:**
Substitute \( \cos A \) and \( \cos B \) into the original equation:
\[ a \left(\frac{b^2 + c^2 - a^2}{2bc}\right) = b \left(\frac{a^2 + c^2 - b^2}{2ac}\right) \]

4. **Clear the denominators by multiplying through by \( 2abc \):**
\[ 2a(b^2 + c^2 - a^2) = 2b(a^2 + c^2 - b^2) \]

5. **Expand both sides:**
\[ 2ab^2 + 2ac^2 - 2a^3 = 2ba^2 + 2bc^2 - 2b^3 \]

6. **Rearrange the equation:**
Move all terms to one side:
\[ 2ab^2 + 2ac^2 - 2a^3 - 2ba^2 - 2bc^2 + 2b^3 = 0 \]

7. **Group similar terms:**
\[ 2ab^2 - 2ba^2 + 2ac^2 - 2bc^2 - 2a^3 + 2b^3 = 0 \]
Factor out the common terms:
\[ 2(a^2b - ab^2 + ac^2 - bc^2 - a^3 + b^3) = 0 \]

8. **Factor the polynomial:**
We can recognize that this can be factored further:
\[ (a^2 - b^2)(c^2 - (a^2 + b^2)) = 0 \]

9. **Set each factor to zero:**
This gives us two cases:
- Case 1: \( a^2 - b^2 = 0 \) which implies \( a = b \) (isosceles triangle).
- Case 2: \( c^2 - (a^2 + b^2) = 0 \) which implies \( c^2 = a^2 + b^2 \) (right triangle).

10. **Conclusion:**
From these cases, we conclude that triangle \( ABC \) is both isosceles (when \( a = b \)) and right-angled (when \( c^2 = a^2 + b^2 \)). Therefore, triangle \( ABC \) is a right-angled isosceles triangle.