If the altitudes of a triangle are in A.P.,th | vedclass

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(B) Let $h_1, h_2, h_3$ be the altitudes from vertices $P, Q, R$ to the opposite sides $a, b, c$ respectively.We know that the area of the triangle $\

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$H.P.$

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(B) Let $h_1, h_2, h_3$ be the altitudes from vertices $P, Q, R$ to the opposite sides $a, b, c$ respectively.We know that the area of the triangle $\Delta = \frac{1}{2} a h_1 = \frac{1}{2} b h_2 = \frac{1}{2} c h_3$.Thus,$h_1 = \frac{2\Delta}{a}$,$h_2 = \frac{2\Delta}{b}$,and $h_3 = \frac{2\Delta}{c}$.Given that $h_1, h_2, h_3$ are in $A.P.$,we have $2h_2 = h_1 + h_3$.Substituting the values,we get $2(\frac{2\Delta}{b}) = \frac{2\Delta}{a} + \frac{2\Delta}{c}$.Dividing by $2\Delta$,we get $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$Therefore,$a, b, c$ are in $H.P.$

If the altitudes of a triangle are in $A.P.$,then the sides of the triangle are in

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