$Let x, y and z be positive real numbers. Suppose x, y and z are lengths of the sides of a triangle$
$opposite to its angles X, Y and Z, respectively. If \tan \frac{X}{2} + \tan \frac{Z}{2} = \frac{2 y}{x + y + z}$
$then which of the following statements is/are TRUE?$

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$2 Y = X + Z$

$x^{2} + z^{2} - y^{2} = xz$

$Y = X + Z$

$\tan \frac{X}{2} = \frac{x}{y + z}$

answer is B, C.

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Detailed Solution

$\begin{array}{l} \tan \frac{x}{2} + \tan \frac{z}{2} = \frac{2 y}{x + y + z} \\ \frac{Δ}{S (S - x)} + \frac{Δ}{S (S - z)} = \frac{2 y}{2 S} \\ \frac{Δ}{S} (\frac{2 S - (x + z)}{(S - x) (S - z)}) = \frac{y}{S} \\ \Rightarrow \frac{Δy}{S (S - x) (S - z)} = \frac{y}{S} \\ \Rightarrow Δ^{2} = (S - x)^{2} (S - z)^{2} \\ \Rightarrow S (S - y) = (S - x) (S - z) \end{array}$

$\begin{array}{l} \Rightarrow (x + y + z) (x + z - y) = (y + z - x) (x + y - z) \\ \Rightarrow (x + z)^{2} - y^{2} = y^{2} - (z - x)^{2} \\ \Rightarrow (x + z)^{2} + (x - z)^{2} = 2 y^{2} \\ \Rightarrow x^{2} + z^{2} = y^{2} \Rightarrow ∠ Y = \frac{π}{2} \\ \Rightarrow ∠ Y = ∠ X + ∠ Z \end{array}$

$\begin{array}{l} \tan \frac{x}{2} = \frac{Δ}{S (S - x)} \\ \tan \frac{x}{2} = \frac{\frac{1}{2} x z}{\frac{(y + z)^{2} - x^{2}}{4}} \\ \tan \frac{x}{2} = \frac{2 x z}{y^{2} + z^{2} + 2 y z - x^{2}} \end{array}$