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(B) Given $	an \frac{X}{2} + 	an \frac{Z}{2} = \frac{2y}{x+y+z}$.Using the formula $	an \frac{X}{2} = \frac{\Delta}{S(S-x)}$ where $S = \frac{x+y+z

Vedclass · Accepted Answer

$B, C$

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(B) Given $	an \frac{X}{2} + 	an \frac{Z}{2} = \frac{2y}{x+y+z}$.Using the formula $	an \frac{X}{2} = \frac{\Delta}{S(S-x)}$ where $S = \frac{x+y+z}{2}$ and $\Delta$ is the area of the triangle,we have:$\frac{\Delta}{S(S-x)} + \frac{\Delta}{S(S-z)} = \frac{2y}{2S} = \frac{y}{S}$$\frac{\Delta}{S} \left( \frac{S-z + S-x}{(S-x)(S-z)} ight) = \frac{y}{S}$$\Delta \left( \frac{2S - (x+z)}{(S-x)(S-z)} ight) = y$Since $2S = x+y+z$,$2S - (x+z) = y$.$\Delta \left( \frac{y}{(S-x)(S-z)} ight) = y \implies \Delta = (S-x)(S-z)$Squaring both sides: $\Delta^2 = (S-x)^2(S-z)^2$$S(S-x)(S-y)(S-z) = (S-x)^2(S-z)^2$$S(S-y) = (S-x)(S-z)$Substituting $S = \frac{x+y+z}{2}$,we get $y^2 = x^2 + z^2$.This implies $\angle Y = 90^\circ$.Since $X+Y+Z = 180^\circ$,$X+Z = 90^\circ$,so $Y = X+Z$. Thus $(B)$ is true.Also,for a right-angled triangle at $Y$,$	an \frac{X}{2} = \sqrt{\frac{(S-y)(S-z)}{S(S-x)}} = \frac{x}{y+z}$. Thus $(C)$ is true.Therefore,$(B)$ and $(C)$ are true.

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