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Answer:
By trigonometric ratio;
XZ/sin(b) = XY/sin(c) = YZ/sin(a)
MN/sin(c) = NZ/sin(f) = MZ/sin(d)
∴ XZ/MZ = XY/MN × (sin(b)/(sin(d))
XY/MN = XZ/MZ × (sin(b)/sin(d))
Where (sin(b)/(sin(d)) = V;
(XZ/MZ)/(XY/MN) = (XY/MN × V)/(XZ/MZ × V)
(XZ/MZ)/(XY/MN) = (XY/MN)/(XZ/MZ)
∴ (XZ/MZ)² = (XY/MN)²
∴ XZ/MZ = XY/MN QED
Step-by-step explanation:
Whereby the shapes in the question are triangles ΔXYZ and ΔMNZ, and given that we have;
XZ/MZ = YZ/NZ, and from the attached drawing, we have;
∠XZY = ∠NZM
By trigonometric ratio, we have;
XZ/sin(b) = XY/sin(c) = YZ/sin(a)
∴ XZ = sin(b)×XY/sin(c)
MN/sin(c) = NZ/sin(f) = MZ/sin(d)
MZ = sin(d)×MN/sin(c)
∴ XZ/MZ = sin(b)×XY/sin(c)/(sin(d)×MN/sin(c)) = XY/MN × (sin(b)/(sin(d))
XY = sin(c) × XZ/sin(b), MN = sin(c) × MZ/sin(d)
XY/MN = sin(c) × XZ/sin(b)/(sin(c) × MZ/sin(d)) = XZ/MZ × (sin(b)/sin(d))
Let 'V' represent (sin(b)/(sin(d)), we have;
XZ/MZ = XY/MN × V...(1)
XY/MN = XZ/MZ × V...(2)
Dividing equation (1) by (2) gives;
(XZ/MZ)/(XY/MN) = (XY/MN × V)/(XZ/MZ × V) = (XY/MN)/(XZ/MZ)
(XZ/MZ)/(XY/MN) = (XY/MN)/(XZ/MZ)
∴ (XZ/MZ)² = (XY/MN)²
∴ XZ/MZ = XY/MN
