Step-by-step Derivations

1. Analyze the given equation. The given relation is: $\tan \frac{X}{2} + \tan \frac{Z}{2} = \frac{2y}{x + y + z}$ Let s be the semi-perimeter of the triangle, so 2s = x + y + z. The right-hand side (RHS) of the equation becomes: $RHS = \frac{2y}{2s} = \frac{y}{s}$

2. Simplify the left-hand side (LHS) using half-angle formulas. We use the formulas for the tangent of half-angles in terms of the area Δ and the semi-perimeter s: $\tan \frac{X}{2} = \frac{Δ}{s(s-x)} \quad \text{and} \quad \tan \frac{Z}{2} = \frac{Δ}{s(s-z)}$ Substituting these into the LHS: $LHS = \frac{Δ}{s(s-x)} + \frac{Δ}{s(s-z)} = \frac{Δ}{s} \left( \frac{1}{s-x} + \frac{1}{s-z} \right)$ $LHS = \frac{Δ}{s} \left( \frac{(s-z) + (s-x)}{(s-x)(s-z)} \right) = \frac{Δ}{s} \left( \frac{2s - x - z}{(s-x)(s-z)} \right)$ Since 2s = x + y + z, we have 2s - x - z = y. Substituting this: $LHS = \frac{Δ}{s} \left( \frac{y}{(s-x)(s-z)} \right)$

3. Equate the LHS and RHS and solve. Now, we equate the simplified expressions for LHS and RHS: $\frac{Δ}{s} \frac{y}{(s-x)(s-z)} = \frac{y}{s}$ Since y > 0 and s > 0, we can cancel the term y/s from both sides: $\frac{Δ}{(s-x)(s-z)} = 1 \implies Δ = (s-x)(s-z)$

4. Use Heron's formula for the area Δ. Heron's formula states $Δ = \sqrt{s(s-x)(s-y)(s-z)}$. Substituting this into our equation: $\sqrt{s(s-x)(s-y)(s-z)} = (s-x)(s-z)$ Squaring both sides (all terms are positive): $s(s-x)(s-y)(s-z) = (s-x)^2(s-z)^2$ Since s > x and s > z in any triangle, s-x > 0 and s-z > 0. We can divide by (s-x)(s-z): $s(s-y) = (s-x)(s-z)$

5. Expand and simplify to find a relation between the sides. $s^2 - sy = s^2 - sx - sz + xz$ $-sy = -s(x+z) + xz$ $s(x+z-y) = xz$ Substitute s = (x+y+z)/2: $\frac{x+y+z}{2} (x+z-y) = xz$ $\frac{((x+z)+y)((x+z)-y)}{2} = xz$ $\frac{(x+z)^2 - y^2}{2} = xz$ $x^2 + 2xz + z^2 - y^2 = 2xz$ $x^2 + z^2 - y^2 = 0 \implies x^2 + z^2 = y^2$

6. Interpret the result. The relation $x^2 + z^2 = y^2$ is the Pythagorean theorem. This means the triangle is a right-angled triangle with the hypotenuse being the side y. Therefore, the angle opposite to side y, which is angle Y, must be a right angle. $Y = 90^\circ \text{ or } \frac{\pi}{2} \text{ radians}$ From the angle sum property of a triangle, $X + Y + Z = 180^\circ$. $X + 90^\circ + Z = 180^\circ \implies X + Z = 90^\circ$

Evaluate the Options

• A: 2Y = X + Z Substituting our findings: $2(90^\circ) = 90^\circ$, which simplifies to $180^\circ = 90^\circ$. This is FALSE.

• B: Y = X + Z Substituting our findings: $90^\circ = X + Z$. This is true from our derivation $X + Z = 90^\circ$. This statement is TRUE.

• C: tan(X/2) = x / (y+z) We can verify this using the properties of the right-angled triangle where $Y = 90^\circ$. We know sin X = x/y and cos X = z/y. Using the half-angle identity for tangent: $\tan \frac{X}{2} = \frac{\sin X}{1 + \cos X} = \frac{x/y}{1 + z/y} = \frac{x/y}{(y+z)/y} = \frac{x}{y+z}$ This statement is TRUE.

• D: $x^2 + z^2 - y^2 = xz$ From the Cosine Rule, we have $\cos Y = \frac{x^2 + z^2 - y^2}{2xz}$. If the statement $x^2 + z^2 - y^2 = xz$ were true, then $\cos Y = \frac{xz}{2xz} = \frac{1}{2}$. This would imply $Y = 60^\circ$, which contradicts our finding that $Y = 90^\circ$. So this statement is FALSE.

Conclusion

The true statements are B and C.