Question

If \(∠\) A  and \(∠\) B are acute angles such that cos A = cos B, then show that
\(∠ A = ∠ B\).

Answer 1

Consider a triangle ABC where CD is perpendicular to AB. Given that cos A = cos B, which implies \( \frac{AD}{AC} = \frac{BD}{BC} \), denoted as equation (1). The objective is to prove that \( ∠A = ∠B \). To achieve this, extend AC to a point P such that BC = CP. From equation (1), we derive \( \frac{AD}{BD} = \frac{AC}{BC} \). This leads to \( \frac{AD}{BD} = \frac{AC}{CP} \) (equation 2). Applying the converse of the Basic Proportionality Theorem, we establish that CD is parallel to BP. Consequently, \( ∠ACD = ∠CPB \) (corresponding angles, equation 3) and \( ∠BCD = ∠CBP \) (alternate interior angles, equation 4). By construction, BC = CP. Therefore, \( ∠CBP = ∠CPB \) (angles opposite equal sides in a triangle, equation 5). Combining equations (3), (4), and (5), we obtain \( ∠ACD = ∠BCD \) (equation 6). Now, consider triangles CAD and CBD. We have \( ∠ACD = ∠BCD \) (from equation 6) and \( ∠CDA = ∠CDB \) (both are 90°). Since two angles are equal, the remaining angles must also be equal. Thus, \( ∠CAD = ∠CBD \), which means \( ∠A = ∠B \).

Alternatively,

Consider a triangle ABC with CD perpendicular to AB. Given that cos A = cos B, it follows that \( \frac{AD}{AC} = \frac{BD}{BC} \). Rearranging this gives \( \frac{AD}{BD} = \frac{AC}{BC} \). Let this common ratio be k, so \( \frac{AD}{BD} = \frac{AC}{BC} = k \). This implies AD = k BD (equation 1) and AC = k BC (equation 2). Using the Pythagorean theorem in triangles CAD and CBD, we get \( (CD)^2 = (AC)^2 - (AD)^2 \) (equation 3) and \( (CD)^2 = (BC)^2 - (BD)^2 \) (equation 4). Equating equations (3) and (4) yields \( (AC)^2 - (AD)^2 = (BC)^2 - (BD)^2 \). Substituting the expressions from equations (1) and (2), we get \( (k BC)^2 - (k BD)^2 = (BC)^2 - (BD)^2 \). Factoring out \( k^2 \) gives \( k^2 (BC^2 - BD^2) = BC^2 - BD^2 \). This simplifies to \( k^2 = 1 \), so k = 1. Substituting k = 1 into equation (2) gives AC = BC. Therefore, \( ∠A = ∠B \) (angles opposite equal sides of a triangle).