Lenghts of the tangents from $A, B, a n d C$ to the incircle are in A.P., then

r_{1}, r_{2}, r_{3}

are in H.P.

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r_{1}, r_{2}, r_{3}

are in A.P.

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a, b, c

are in A.P.

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cos A = \frac{4 c - 3 b}{2 b}

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Solution

The correct options are
B $r_{1}, r_{2}, r_{3}$ are in H.P.
C $a, b, c$ are in A.P.
From the figure:
$A F = A E = a ‘$ ; $B F = B D = b ‘$ ; $C D = C E = c ‘$
$c = A B = A F + F B = a ‘ + b ‘$ ; $a = B C = B D + D C = b ‘ + c ‘$ ; $a = B C = B D + D C = b ‘ + c ‘$
given that, $A F, B D, C E$ are in A.P
$\Rightarrow 2 B D = A F + C E$
$\Rightarrow 2 b ‘ = a ‘ + c ‘$
$\Rightarrow (b ‘ + c ‘) + (a ‘ + b ‘) = 2 (a ‘ + c ‘)$
$\Rightarrow a + c = 2 b$
Therefore, $a, b, c$ are in A.P
$\Rightarrow \frac{s - a}{△}, \frac{s - b}{△} . \frac{s - c}{△}$ are in A.P
$\Rightarrow \frac{1}{r_{1}}, \frac{1}{r_{2}}, \frac{1}{r_{3}}$ are in A.P
$\Rightarrow r_{1}, r_{2}, r_{3}$ are in H.P
Ans: A,B,C

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