91Ó°ÊÓ

Let's consider a triangle ABC, where its angles are A, B, and C, and its sides opposite to these angles are a, b, and c respectively. Let h_a, h_b, and h_c be the altitudes from vertices A, B, and C respectively. Using the formula for the area of a triangle, we can represent the altitudes in terms of the respective angles. Area of triangle ABC can be given by three equivalent expressions using the altitudes: \(Area = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B = \frac{1}{2} ab \sin C\) Now we can represent the altitudes h_a, h_b, and h_c as follows: \(h_a = \frac{2*Area}{b*c} = \frac{\sin A}{\sin B * \sin C}\) \(h_b = \frac{2*Area}{a*c} = \frac{\sin B}{\sin A * \sin C}\) \(h_c = \frac{2*Area}{a*b} = \frac{\sin C}{\sin A * \sin B}\)

Given that \(\sin A\), \(\sin B\), and \(\sin C\) are in A.P., which means there exists a common difference 'd' such that: \(\sin B = \sin A + d\) \(\sin C = \sin B + d = \sin A + 2d\)

To prove that the altitudes h_a, h_b, and h_c are in H.P., we need to show that their reciprocals are in A.P. Let's find the reciprocals of the altitudes. \(h_a^{-1} = \frac{\sin B * \sin C}{\sin A}\) \(h_b^{-1} = \frac{\sin A * \sin C}{\sin B}\) \(h_c^{-1} = \frac{\sin A * \sin B}{\sin C}\) Now, let's use the expressions from step 2: \(h_a^{-1} = \frac{(\sin A + d) * (\sin A + 2d)}{\sin A}\) \(h_b^{-1} = \frac{\sin A * (\sin A + 2d)}{(\sin A + d)}\) \(h_c^{-1} = \frac{\sin A * (\sin A + d)}{(\sin A + 2d)}\) From the above expressions, we can find the common difference 'd' between the reciprocals of altitudes: \(d_1 = h_b^{-1} - h_a^{-1}= \frac{\sin A * (\sin A + 2d)}{(\sin A + d)} - \frac{(\sin A + d) * (\sin A + 2d)}{\sin A}\) \(d_2 = h_c^{-1} - h_b^{-1}= \frac{\sin A * (\sin A + d)}{(\sin A + 2d)} - \frac{\sin A * (\sin A + 2d)}{(\sin A + d)}\) By simplifying the expressions for \(d_1\) and \(d_2\), we can see that they are equal, which means the reciprocals of altitudes are in A.P.: \(d_1 = d_2 = \frac{2d\sin A}{(\sin A + d)}\) Thus, the altitudes \(h_a\), \(h_b\), and \(h_c\) are in Harmonic Progression (H.P.).